Maths M.C.Q (1 Marks)

$ \Rightarrow $ $8 = \frac{2}{{r(1 - r)}}\left( {\;a = \frac{2}{r}} \right)$

$ \Rightarrow $ $4r(1 - r) = 1 $

$\Rightarrow 4r - 4{r^2} - 1 = 0$

$ \Rightarrow $ $4{r^2} - 4r + 1 = 0$

$\Rightarrow \left( {r - \frac{1}{2}} \right)(4r - 2) = 0$

$\Rightarrow r = \frac{1}{2}$

So first term $a = 4$.

$x = \frac{1}{{1 - a}}$

$\Rightarrow a = \frac{{x - 1}}{x}$ and $y = \frac{1}{{1 - b}}$

$\Rightarrow b = \frac{{y - 1}}{y}$

$\therefore $$1 + ab + {a^2}{b^2} + ..........\infty = \frac{1}{{1 - ab}}$

$ = \frac{1}{{1 - \frac{{x - 1}}{x}.\frac{{y - 1}}{y}}} = \frac{{xy}}{{x + y - 1}}$.

$A = 1 + [{r^z} + {r^{2z}} + {r^{3z}} + ........\infty ]$

We know that sum of infinite $G.P.$ is

${S_\infty } = \frac{a}{{1 - r}}( - 1 < r < 1)$

Therefore, $A = 1 + \left[ {\frac{{{r^z}}}{{1 - {r^z}}}} \right]$

$\Rightarrow A = \frac{{1 - {r^z} + {r^z}}}{{1 - {r^z}}}$

$\therefore $ $A = \frac{1}{{1 - {r^z}}}$

$\Rightarrow 1 - {r^z} = \frac{1}{A} $

$\Rightarrow {r^z} = \frac{{A - 1}}{A}$

Hence $r = {\left[ {\frac{{A - 1}}{A}} \right]^{1/z}}$.

Under conditions, we get $\frac{a}{r}\;.\;a\;.\;ar = 216$

$ \Rightarrow $ $a = 6$

And sum of product pair wise $ = 156$

$ \Rightarrow $ $\frac{a}{r}\;.\;a + \frac{a}{r}\;.\;ar + a\;.\;ar = 156$

$ \Rightarrow $ $r = 3$

Hence numbers are $2, 6, 18.$

Trick : Since $2 \times 6 \times 18 = 216$ (as given) and no other option gives the value.

Given series can be written as,

$ = \frac{1}{3}[9 + 99 + 999 + ........ + n\,\,{\rm{terms]}}$

$ = \frac{1}{3}\left[ {(10 - 1) + ({{10}^2} - 1) + ({{10}^3} - 1) + .... + n\,{\rm{terms}}} \right]$

$ = \frac{1}{3}\left[ {10 + {{10}^2} + .... + {{10}^n}} \right]$$ - \frac{1}{3}\left[ {1 + 1 + 1 + .... + n\,{\rm{terms}}} \right]$

$ = \frac{1}{3}\,.\,\frac{{10\,({{10}^n} - 1)}}{{10 - 1}} - \frac{1}{3}.n\,$ $ = \frac{1}{3}\left[ {\frac{{{{10}^{n + 1}} - 10}}{9} - n} \right]$

$ = \frac{1}{3}\,\left[ {\frac{{{{10}^{n\, + \,1}} - 9n - 10}}{9}} \right]$ $ = \frac{1}{{27}}[{10^{n\, + \,1}} - 9n - 10]$.

$ = \frac{{1.({{10}^{91}} - 1)}}{{10 - 1}} = \frac{{{{({{10}^{13}})}^7} - 1}}{{{{10}^{13}} - 1}} \times \frac{{{{10}^{13}} - 1}}{{10 - 1}}$

$ = [{({10^{13}})^6} + {({10^{13}})^5} + {({10^{13}})^4} + ......1]$

$({10^{12}} + {10^{11}} + ...... + 1)$

It is the product of two integers and hence not prime.

$a{r^{n - 1}} = 128$ …..$(ii)$

and common ratio $r = 2$ …..$(iii)$

From $(iii), (i)$ and $(ii)$

we get $a{2^{n - 1}} = 128$  …..$(iv)$

and $\frac{{a({2^n} - 1)}}{{2 - 1}} = 255$ .....$(v)$

Dividing $(v)$ by $(iv)$

we get $\frac{{{2^n} - 1}}{{{2^{n - 1}}}} = \frac{{255}}{{128}}$

$ \Rightarrow $$2 - {2^{ - n + 1}} = \frac{{255}}{{128}}$

$ \Rightarrow $${2^{ - n}} = {2^{ - 8}}$

$ \Rightarrow $$n = 8$

Putting $n = 8$ in equation $(iv),$

we have $a\;.\;{2^7} = 128 = {2^7}$or $a = 1$.

then under condition,

${T_n} = {T_{n - 1}} + {T_{n - 2}}$

$ \Rightarrow $ $a{r^{n - 1}} = a{r^{n - 2}} + a{r^{n - 3}}$

$ \Rightarrow $ $a{r^{n - 1}} = a{r^{n - 1}}{r^{ - 1}} + a{r^{n - 1}}{r^{ - 2}}$

$ \Rightarrow $ $1 = \frac{1}{r} + \frac{1}{{{r^2}}}$

$ \Rightarrow $ ${r^2} - r - 1 = 0$

$ \Rightarrow $ $r = \frac{{1 \pm \sqrt {1 + 4} }}{2} = \frac{{1 + \sqrt 5 }}{2}$

Taking only $(+)$ sign . $(\because \;r > 1)$

= $0.2 + 0.034 + 0.00034 + 0.0000034 +.............$                                         

= $0.2 + \frac{{34}}{{1000}} + \frac{{34}}{{100000}} + \frac{{34}}{{10000000}} + .....\infty $

$ = \frac{2}{{10}} + 34\left[ {\frac{1}{{{{10}^3}}} + \frac{1}{{{{10}^5}}} + \frac{1}{{{{10}^7}}} + ........\infty } \right]$

$ = \frac{2}{{10}} + 34\left[ {\frac{{1/{{10}^3}}}{{1 - 1/1000}}} \right] = \frac{2}{{10}} + 34 \times \frac{1}{{1000}} \times \frac{{100}}{{99}}$

$ = \frac{2}{{10}} + \frac{{34}}{{990}} = \frac{{232}}{{990}}$.

$\therefore $${S_{100}} = a\left( {\frac{{1 - {r^{100}}}}{{1 - r}}} \right) = \frac{9}{{10}}\left( {\frac{{1 - \frac{1}{{{{10}^{100}}}}}}{{1 - \frac{1}{{10}}}}} \right) = 1 - \frac{1}{{{{10}^{100}}}}$.

$ \Rightarrow a{r^{n - 1}} = a{r^n} + a{r^{n + 1}} $

$\Rightarrow {r^{n - 1}} = {r^n}(1 + r)$

$ \Rightarrow $ ${r^2} + r - 1 = 0$

$ \Rightarrow r = \frac{{ - 1 \pm \sqrt {1 + 4} }}{2} = \frac{{ - 1 \pm \sqrt 5 }}{2}$

Since, each term is $ + ve$.

Hence common ratio is $\frac{{\sqrt 5 - 1}}{2}$.

${r^{p + q - 1 - p + q + 1}} = \frac{m}{n}$

$\Rightarrow r = {\left( {\frac{m}{n}} \right)^{1/(2q)}}$

and $a = \frac{m}{{{{\left( {\frac{m}{n}} \right)}^{(p + q - 1)/2q}}}}$

Now ${p^{th}}$ term $ = a{r^{p - 1}} = \frac{m}{{{{\left( {\frac{m}{n}} \right)}^{(p + q - 1)/2q}}}}{\left( {\frac{m}{n}} \right)^{(p - 1)/2q}}$

$ = m{\left( {\frac{m}{n}} \right)^{(p - 1)/2q - (p + q - 1)/(2q)}} = m{\left( {\frac{m}{n}} \right)^{ - 1/2}} = {m^{1 - 1/2}}{n^{1/2}}$

$ = {m^{1/2}}{n^{1/2}} = \sqrt {mn} $.

Aliter : As we know each term in a $G.P.$ is geometric mean of the terms equidistant from it.

Here ${(p + q)^{th}}$and ${(p - q)^{th}}$ terms are equidistant from ${p^{th}}$ term

$i.e.$ at a distance of $q$.

Therefore, ${p^{th}}$ term will be $G.M.$ of ${(p + q)^{th}}$ and ${(p - q)^{th}}$

$i.e.$ $\sqrt {mn} $.

$ \Rightarrow $ ${b^2}(a - c) = ac(a - c)$

$ \Rightarrow $${b^2}a - {b^2}c = {a^2}c - a{c^2}$

$ \Rightarrow $ $a({b^2} + {c^2}) = c({a^2} + {b^2})$.

Trick : Put $a = 1,\;b = 2,\;c = 4$ and check the alternates.

==> ${({a_1}.{a_2}.{a_3})^{1/3}} \ge \frac{3}{{(1/{a_1} + 1/{a_2} + 1/{a_3})}}$

==> $({a_1}.\,{a_2}.{a_3})\, \ge \,\frac{{27}}{{{{\left( {1/{a_1} + 1/{a_2} + 1/{a_3}} \right)}^3}}}$

$({a_1}.\,{a_2}.{a_3})\,{\left( {\frac{1}{{{a_1}}} + \frac{1}{{{a_2}}} + \frac{1}{{{a_3}}}} \right)^3}$$ \ge 27$.

${t_1} = \log 25$; ${s_1} = {\left[ {\log \,\,5/3} \right]^1}$

${t_2} = \log \frac{{125}}{3}$; ${s_2} = {\left[ {\log \,5/3} \right]^2}$

${t_3} = \log \frac{{625}}{9}$; ${s_3} = {\left[ {\log \,5/3} \right]^3}$

Clearly ${t_n}$ is an $A.P.$ and ${s_n}$ is $G.P.$

$a + ar + a{r^2} = 14$$ \Rightarrow a\,(1 + r + {r^2}) = 14$ …..$(i)$

and $2\,(ar + 1) = (a + 1) + (a{r^2} - 1)$

$a\,({r^2} - 2r + 1) = 2$ ….. $(ii)$

Put the value of a from $(i)$ to $(ii),$

==> $2{r^2} - 5r + 2 = 0$

==> $r = 2,\frac{1}{2}\,\,{\rm{and}}\,\,a = 2,8$

$\therefore $ Numbers are $2, 4, 8$ or $8, 4, 2$. So lowest term in series is $2$.

Now $G_1^3 + G_2^3 = 72$.

Also option $(b)$ gives this value

$i.e.$  $2 \times 2 \times 4 \times \frac{9}{2} = 72$.

if $\alpha ,\;\beta $ are roots of quadratic equation, then quadratic equation is

${x^2} - x(\alpha + \beta ) + \alpha \beta = 0$ ......$(i)$

$A.M. = \frac{{\alpha + \beta }}{2} = 8$

$ \Rightarrow $$\alpha + \beta = 16$ ......$(ii)$

and $G.M. = \sqrt {\alpha \beta } = 5$

$ \Rightarrow $$\alpha \beta = 25$ ......$(iii)$

So the required quadratic equation will be ${x^2} - 16x + 25 = 0$.

Now $A.M.$$ = \frac{{a + b}}{2}$

and $G.M.$$ = \sqrt {ab} $

Under conditions,

$A.M.=G.M.+ 2$

$ \Rightarrow $$\frac{{a + b}}{2} = \sqrt {ab} + 2$ ......$(i)$

and $\frac{a}{b} = \frac{4}{1}$

$ \Rightarrow $$a = 4b$ .....$(ii)$

From $(ii)$ and $(i),$ we get $a = 16$ and $b = 4$.

Condition $I$ : $\frac{a}{r} + a + ar = 14 $

$\Rightarrow a\left( {\frac{1}{r} + 1 + r} \right) = 14$..(i)

Condition $II$ : $\frac{a}{r} + 1,\;a + 1$ and $ar - 1$ will be in $A.P.$,

then $2(a + 1) = \frac{a}{r} + 1 + ar - 1 = \frac{a}{r}(1 + {r^2})$ ......(ii)

From (i) and (ii), we get $a = 4$ and $r = 2$.

So, required numbers are $2, 4, 8.$

Hence greatest number is $8.$

$\frac{{a + b}}{2} > \sqrt {ab} ,\;\frac{{b + c}}{2} > \sqrt {bc} $

and $\frac{{a + c}}{2} > \sqrt {ac} $.

Multiplying these inequalities, we get

$(a + b)(b + c)(c + a) > 8abc$.

$\frac{{x + y}}{{2(\sqrt {xy} )}} = \frac{p}{q}$…..$(i)$

$\frac{{{x^2} + {y^2} + 2xy}}{{4xy}} = \frac{{{p^2}}}{{{q^2}}}$

$\frac{{{x^2} + {y^2} + 2xy - 4xy}}{{4xy}} = \frac{{{p^2} - {q^2}}}{{{q^2}}}$

$\frac{{{{(x - y)}^2}}}{{4xy}} = \frac{{{p^2} - {q^2}}}{{{q^2}}}$

$\frac{{x - y}}{{2\sqrt {xy} }} = \frac{{\sqrt {{p^2} - {q^2}} }}{q}$…..$(ii)$

Equation $(i)$ is divided by $(ii),$

Then $\frac{{x + y}}{{x - y}} = \frac{p}{{\sqrt {{p^2} - {q^2}} }}$;

$\frac{x}{y} = \frac{{p + \sqrt {{p^2} - {q^2}} }}{{p - \sqrt {{p^2} - {q^2}} }}$.

$\therefore {T_2} = a + d,\,\,$

${T_{10}} = a + 9d,$

${T_{34}} = a + 33d$

$\therefore {(a + 9d)^2} = (a + d)(a + 33d)$

==> ${a^2} + 81{d^2} + 18ad = {a^2} + ad + 33ad + 33{d^2}$

Put $a = 1$

$ \Rightarrow 1 + 81{d^2} + 18d = 1 + d + 33d + 33{d^2}$

==> $48{d^2} - 16d = 0$

$ \Rightarrow 16d(3d - 1) = 0$

==> $d = 0,\,\,d = 1/3$.

$a + d + a + a - d = 27$

==> $3a = 27$

==> $a = 9$

Now, $(a + d - 1),(a - 1),(a - d + 3)$are in $G.P.$

$ \Rightarrow $${(a - 1)^2} = (a + d - 1)(a - d + 3)$

$ \Rightarrow 64 = (8 + d)(12 - d)$

$ \Rightarrow 64 = - {d^2} + 4d + 96$

$ \Rightarrow {d^2} - 4d - 32 = 0$

$ \Rightarrow {d^2} - 8d + 4d - 32 = 0$

==> $(d - 8)(d + 4) = 0$,

$\therefore d = - 4,\,8$

Series is $5, 9, 13$ (for $d = -4$) and $17, 9, 1$ (for $d = 8$)

Decreasing $A.P.$ is $17, 9, 1.$

Then ${(a - d)^2},\;{a^2},\;{(a + d)^2}$ are in $G.P.$

$\therefore $${a^4} = {(a - d)^2}{(a + d)^2}$

$ \Rightarrow $${d^4} - 2{a^2}{d^2} = 0$

$ \Rightarrow $$d = 0,\; \pm \sqrt 2 a$

Hence $d$ has three values.

So $\frac{a}{r}.\;a.\;ar = 512$

$ \Rightarrow $${a^3} = {8^3}$

$ \Rightarrow $$a = 8$

From second condition, we get $\frac{a}{r} + 8,\;a + 6,\;ar$ will be in $A.P.$

$ \Rightarrow $ $2(a + 6) = \frac{a}{r} + 8 + ar$

$ \Rightarrow $$28 = 8\left\{ {\frac{1}{r} + 1 + r} \right\}$

$ \Rightarrow $ $\frac{1}{r} + r + 1 = \frac{7}{2}$

$ \Rightarrow $$\frac{1}{r} + r - \frac{5}{2} = 0$

$ \Rightarrow $ ${r^2} - \frac{5}{2}r + 1 = 0$

$ \Rightarrow $$2{r^2} - 5r + 2 = 0$

$ \Rightarrow $ $(2r - 1)(r - 2) = 0$

$ \Rightarrow $$r = \frac{1}{2},\;r = 2$ $(\because \;r > 1)$

$ \Rightarrow $ $r = 2$.

Hence required numbers are $4,\;8,\;16$.

Trick : Check for $(a)$ $2 + 8,\;4 + 6,\;8$ are not in $A.P.$

$(b)$ $4 + 8,\;8 + 6,\;16\;i.e.\;12,\;14,\;16$ are in $A.P.$

Applying componendo and dividendo, we get

$\frac{{2a}}{{2bx}} = \frac{{2b}}{{2cx}} = \frac{{2c}}{{2dx}}$

$ \Rightarrow $${b^2} = ac$ and ${c^2} = bd$

$ \Rightarrow $$a,\;b,\;c$ and $b,\;c,\;d$ are in $G.P.$

Therefore, $a,\;b,\;c,\;d$ are in $G.P.$

$ \Rightarrow $${({\log _z}x)^2} = {\log _x}y \times {\log _y}z = {\log _x}z = \frac{1}{{{{\log }_z}x}}$

$ \Rightarrow $${({\log _z}x)^3} = 1$

$ \Rightarrow $$z = x$

Also, we can show $z = x = y = 4$.

or $\frac{1}{2}(a + b) = 2\sqrt {ab} $

or $\frac{{a + b}}{{2\sqrt {ab} }} = \frac{2}{1}$

$ \Rightarrow $$\frac{{a + b + 2\sqrt {ab} }}{{a + b - 2\sqrt {ab} }} = \frac{{2 + 1}}{{2 - 1}} = \frac{3}{1}$

$ \Rightarrow $ $\frac{{{{(\sqrt a + \sqrt b )}^2}}}{{{{(\sqrt a - \sqrt b )}^2}}} = \frac{3}{1}$

$ \Rightarrow $ $\frac{{\sqrt a + \sqrt b }}{{\sqrt a - \sqrt b }} = \frac{{\sqrt 3 }}{1}$

$ \Rightarrow $ $\frac{a}{b} = {\left( {\frac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}}} \right)^2}$

$ \Rightarrow $ $\frac{a}{b} = \frac{{2 + \sqrt 3 }}{{2 - \sqrt 3 }}$

or $a:b = (2 + \sqrt 3 ):(2 - \sqrt 3 )$.

So, ${b^2} = ac$…..(i)

$x = \frac{{a + b}}{2}$…..(ii)

$y = \frac{{b + c}}{2}$…..(iii)

Now $\frac{a}{x} + \frac{c}{y} = \frac{{2a}}{{a + b}} + \frac{{2c}}{{b + c}} = \frac{{2(ab + bc + 2ca)}}{{ab + ac + {b^2} + bc}}$

$ = \frac{{2(ab + bc + 2ca)}}{{(ab + ac + ac + bc)}} = 2$,$\left\{ {\because \;{b^2} = ac} \right\}$.

Trick : Let $a = 1,\;b = 2,\;c = 4,$ then obviously $x = \frac{3}{2}$

and $y = 3$, then $\frac{1}{{3/2}} + \frac{4}{3} = 2$.

and geometric mean $G = \sqrt {ab} $

Then $A - G = \frac{{a + b}}{2} - \sqrt {ab} $$ = \frac{{a + b - 2\sqrt {ab} }}{2}$

$ = \frac{{{{(\sqrt a )}^2} + {{(\sqrt b )}^2} - 2(\sqrt a )(\sqrt b )}}{2} = {\left[ {\frac{{\sqrt a - \sqrt b }}{{\sqrt 2 }}} \right]^2}$

${\rm{A}}{\rm{.M}}{\rm{.}}\, \ge {\rm{G}}{\rm{.M}}{\rm{.}}$

$\left( {\frac{{{x_1} + {x_2} + {x_3} + ...... + {x_n}}}{n}} \right)\, \ge \,{({x_1}.{x_2}.{x_3}.......{x_n})^{\frac{1}{n}}}$

$ = {(1)^{\frac{1}{n}}} = 1$

${x_1} + {x_2} + {x_3} + ........ + {x_n} \ge n$

${x_1} + {x_2} + {x_3} + ....... + {x_n}$ can never be less than $n$.

${T_n} = n{(2n + 1)^2} = 4{n^3} + 4{n^2} + n$

$\therefore $ ${S_n} = \sum\limits_1^n {{T_n}} = \sum\limits_1^n {(4{n^3} + 4{n^2} + n)} $

$ = 4\sum\limits_1^n {{n^3}} + 4\sum\limits_1^n {{n^2}} + \sum\limits_1^n n $

$ = 4{\left\{ {\frac{n}{2}(n + 1)} \right\}^2} + \frac{4}{6}n(n + 1)(2n + 1) + \frac{n}{2}(n + 1)$

$ = \frac{n}{6}(n + 1)(6{n^2} + 14n + 7)$.

$\frac{1}{5}{S_n} = {\rm{ }}\frac{1}{5} + \frac{2}{{{5^2}}} + \frac{3}{{{5^3}}} + ....... + \frac{n}{{{5^n}}}$

Subtracting,

$\left( {1 - \frac{1}{5}} \right){S_n} = 1 + \frac{1}{5} + \frac{1}{{{5^2}}} + \frac{1}{{{5^3}}} + ...... + {\rm{upto}}\;n\;{\rm{terms}}\; - \frac{n}{{{5^n}}}$

$ \Rightarrow $$\frac{4}{5}{S_n} = \frac{{1 - \frac{1}{{{5^n}}}}}{{\frac{4}{5}}} - \frac{n}{{{5^n}}}$

$ \Rightarrow $${S_n} = \frac{{25}}{{16}} - \frac{{4n + 5}}{{16 \times {5^{n - 1}}}}$.

$S = 2 + 4 + 7 + 11 + 16 + ........{T_{n - 1}} + {T_n}$

Subtracting, we get

$0 = 2 + \left\{ {2 + 3 + 4 + ........ + ({T_n} - {T_{n - 1}})} \right\} - {T_n}$

$ \Rightarrow $${T_n} = 1 + (1 + 2 + 3 + 4 + ......{\rm{upto}}\;n\;{\rm{terms}})$

$ \Rightarrow $$1 + \frac{1}{2}n(n + 1) = \frac{{2 + {n^2} + n}}{2} $

$= \frac{{{n^2} + n + 2}}{2}$.

${S_n} = 12 + 16 + 24 + 40 + ..... + {T_n}$

Again ${S_n} = \,12 + 16 + 24 + ...... + {T_n}$

On subtraction

$0 = (12 + 4 + 8 + 16 + ...$+ upto $n$ terms) -${T_n}$

or ${T_n} = 12 + [4 + 8 + 16 + ... + {\rm{upto }}(n - 1)$ terms]

$ = 12 + \frac{{4({2^{n - 1}} - 1)}}{{2 - 1}} = {2^{n + 1}} + 8$

On putting $n = 1,\,2,\,3......$

${T_1} = {2^2} + 8$, ${T_2} = {2^3} + 8$, ${T_3} = {2^4} + 8......etc.$

${S_n} = {T_1} + {T_2} + {T_3} + .... + {T_n}$

$ = ({2^2} + {2^3} + {2^4} + ....{\rm{upto}}\,\,\,n\,\,\,{\rm{terms)}}$

$ + (8 + 8 + 8 + ......$upto $n$ terms)

$ = \frac{{{2^2}({2^n} - 1)}}{{2 - 1}} + 8n = 4({2^n} - 1) + 8n.$

Again $S = {\rm{ }}2 + 4 + 7 + 11 + ....... + {T_{n - 1}} + {T_n}$

Subtracting, we get

$0 = 2 + \left\{ {2 + 3 + 4 + 5 + .....({T_n} - {T_{n - 1}})} \right\} - {T_n}$

${T_n} = 2 + \frac{1}{2}(n - 1)(4 + \{ n - 2)1\} = \frac{1}{2}({n^2} + n + 2)$

Now $S = \Sigma {T_n} = \frac{1}{2}\Sigma ({n^2} + n + 2) $

$= \frac{1}{2}(\Sigma {n^2} + \Sigma n + 2\Sigma \,1)$

$ = \frac{1}{2}\left\{ {\frac{1}{6}n(n + 1)(2n + 1) + \frac{1}{2}n(n + 1) + 2n} \right\}$

$ = \frac{n}{{12}}\left\{ {(n + 1)(2n + 1 + 3) + 12} \right\}$

= $\frac{n}{6}\left\{ {(n + 1)(n + 2) + 6} \right\} $

$= \frac{n}{6}({n^2} + 3n + 8)$.

$ \Rightarrow $$x.S = x + 3{x^2} + 6{x^3} + .......\infty $

Subtracting $S(1 - x) = 1 + 2x + 3{x^2} + 4{x^3} + .......\infty $

$ \Rightarrow $$x(1 - x)S = x + 2{x^2} + 3{x^3} + .......\infty $

Again subtracting,

$ \Rightarrow $$S[(1 - x) - x(1 - x)] = 1 + x + {x^2} + {x^3} + ........\infty $

$ \Rightarrow $$S[(1 - x)(1 - x)] = \frac{1}{{1 - x}} $

$\Rightarrow S = \frac{1}{{{{(1 - x)}^3}}}$

$ \Rightarrow $ $\frac{1}{5}S = {\rm{ }}\frac{1}{5} + 4.\frac{1}{{{5^2}}} + 7.\frac{1}{{{5^3}}} + .........$

Subtracting $\left( {1 - \frac{1}{5}} \right)S = 1 + 3.\frac{1}{5} + 3.\frac{1}{{{5^2}}} + 3.\frac{1}{{{5^3}}} + ........$

$ = 1 + 3\left( {\frac{1}{5} + \frac{1}{{{5^2}}} + ......} \right)$

$ \Rightarrow $$\frac{4}{5}.S = 1 + 3.\frac{1}{5}\left( {\frac{1}{{1 - \frac{1}{5}}}} \right) = 1 + \frac{3}{4} = \frac{7}{4} \Rightarrow S = \frac{{35}}{{16}}$.

Aliter : Use direct formula ${S_\infty } = \frac{{ab}}{{1 - r}} + \frac{{dbr}}{{{{(1 - r)}^2}}}$

Here $a = 1,\;b = 1,\;d = 3,\;r = \frac{1}{5}$, therefore

${S_\infty } = \frac{1}{{1 - \frac{1}{5}}} + \frac{{3 \times 1 \times \frac{1}{5}}}{{{{\left( {1 - \frac{1}{5}} \right)}^2}}} = \frac{5}{4} + \frac{{\frac{3}{5}}}{{\frac{{16}}{{25}}}} = \frac{5}{4} + \frac{{15}}{{16}} = \frac{{35}}{{16}}$.

Aliter : Use $S = \left[ {1 + \frac{r}{{1 - r}} \times {\rm{diff}}{\rm{.}}\;{\rm{of}}\;{\rm{A}}{\rm{.P}}{\rm{.}}} \right]\frac{1}{{1 - r}}$

Hence sum of $2n$ terms is
$n$.

Let $s = 1 - \alpha \beta + {\alpha ^2}{\beta ^2}.......$

==> $s = \frac{1}{{1 + \alpha \beta }}$
$\alpha = \frac{1}{{{s_1}}} - 1,\,\,\beta = \frac{1}{{{s_2}}} - 1$;

$\therefore s = \frac{1}{{1 + \left( {\frac{1}{{{s_1}}} - 1} \right)\,\left( {\frac{1}{{{s_2}}} - 1} \right)}}$.

$s = \frac{{{s_1}{s_2}}}{{2{s_1}{s_2} + 1 - {s_1} - {s_2}}}$.

Rationalization of ${D^r}$

$\therefore S = (\sqrt 2 - \sqrt 1 ) + \left( {\sqrt 3 - \sqrt 2 } \right) + ... + \left( {\sqrt {{n^2}} - \sqrt {{n^2} - 1} } \right)$

$S = n -1.$
 

$= \frac{{\frac{{n(n + 1)}}{4}}}{{{{\left( {\frac{{n(n + 1)}}{2}} \right)}^2}}} $

$= \frac{1}{{n(n + 1)}} = \frac{1}{n} - \frac{1}{{n + 1}}$

${S_n} = \sum\limits_{}^{} {\left( {\frac{1}{n} - \frac{1}{{n + 1}}} \right)} $

$ = \left( {1 - \frac{1}{2}} \right) + \left( {\frac{1}{2} - \frac{1}{3}} \right) + \left( {\frac{1}{3} - \frac{1}{4}} \right) + ....... + \left( {\frac{1}{n} - \frac{1}{{n + 1}}} \right)$

$ = 1 - \frac{1}{{n + 1}} = \frac{n}{{n + 1}}$.

$S = [2n + 2.2n + 3.2n + ...... + n.2n] - $

$[1 + 2.3 + 3.5 + .... + n.(2n - 1)]$

Let,${S_1} = 2n(1 + 2 + 3 + .... + n)$

= $\frac{{2n.n(n + 1)}}{2} = {n^2}(n + 1)$

and ${S_2} = 1 + 2.3 + 3.5 + ..... + n.(2n - 1)$

${T_n} = n(2n - 1) = 2{n^2} - n$

$\therefore {S_2} = \sum (2{n^2} - n)$

$ = 2\sum ({n^2}) - \sum (n)$

$ = \frac{{2n(n + 1)(2n + 1)}}{6} - \frac{{n(n + 1)}}{2}$

$\therefore S = {S_1} - {S_2} = {n^2}(n + 1) - \frac{{2n(n + 1)(2n + 1)}}{6} + \frac{{n(n + 1)}}{2}$

$ = n\,(n + 1)\left[ {n - \frac{{2n + 1}}{3} + \frac{1}{2}} \right]$

$ = n\,(n + 1)\left[ {\frac{{6n - 4n - 2 + 3}}{6}} \right]$

$ = \frac{{n\,(n + 1)(2n + 1)}}{6}$.

$= \left[ {\frac{{n(n + 1)}}{2}} \right]_{n = 20}^2 - \left[ {\frac{{n(n + 1)}}{2}} \right]_{n = 10}^2$

$= 44100 -3025 = 41075.$

and second factors of the terms of the given series $2,\;3,\;4,,\;........(n + 1)$

${n^{th}}$term of the given series $ = n(n + 1) = {n^2} + n$

Hence sum = $\Sigma {n^2} + \Sigma n = \frac{1}{6}n(n + 1)(2n + 1) + \frac{n}{2}(n + 1)$

$ = \frac{1}{6}n(n + 1)(2n + 1 + 3) = \frac{1}{3}n(n + 1)(n + 2)$.

${S_n} = n - \sum\limits_{n = 1}^n {{{\left( {\frac{1}{3}} \right)}^n}} $

$ = n - \frac{{\frac{1}{3}\left[ {1 - {{\left( {\frac{1}{3}} \right)}^n}} \right]}}{{\left( {1 - \frac{1}{3}} \right)}}$

$ = n - \frac{1}{2}(1 - {3^{ - n}}) = n + \frac{1}{2}({3^{ - n}} - 1)$.

$1\;.\;3\;.\;5 + 2\;.\;5\;.\;8 + 3\;.\;7\;.\;11 + ....... + n(2n + 1)(3n + 2)\;$

So, ${T_n} = n(2n + 1)(3n + 2) = n\,[6{n^2} + 4n + 3n + 2]$

${T_n} = 6{n^3} + 7{n^2} + 2n$

Now, sum $ = 6\Sigma {n^3} + 7\Sigma {n^2} + 2\Sigma n$

$ = 6{\left[ {\frac{1}{2}n(n + 1)} \right]^2} + 7\left[ {\frac{1}{6}n(n + 1)(2n + 1)} \right] + 2\left[ {\frac{1}{2}n(n + 1)} \right]$

$ = \frac{1}{6}n(n + 1)[9n\,(n + 1) + 7(2n + 1) + 6]$

$ = \frac{1}{6}n\,(n + 1)[9{n^2} + 9n + 14n + 7 + 6]$

$ = \frac{{n\,(n + 1)(9{n^2} + 23n + 13)}}{6}$.

${S_n} = \Sigma ({T_n}) = \Sigma \,6\,\left[ {\frac{1}{n} - \frac{1}{{n + 1}}} \right]$$ = 6\left[ {1 - \frac{1}{{n + 1}}} \right]$

${S_n} = \frac{{6n}}{{n + 1}}$.