(Each question 4 marks)

Question 14 Marks

Show that $\frac { 1 \times 2 ^ { 2 } + 2 \times 3 ^ { 2 } + \ldots \ldots + n \times ( n + 1 ) ^ { 2 } } { 1 ^ { 2 } \times 2 + 2 ^ { 2 } \times 3 + \ldots \ldots + n ^ { 2 } ( n + 1 ) } = \frac { 3 n + 5 } { 3 n + 1 }$

Answer

Given: $\frac { 1 \times 2 ^ { 2 } + 2 \times 3 ^ { 2 } + \ldots \ldots + n \times ( n + 1 ) ^ { 2 } } { 1 ^ { 2 } \times 2 + 2 ^ { 2 } \times 3 + \ldots \ldots + n ^ { 2 } ( n + 1 ) } = \frac { 3 n + 5 } { 3 n + 1 }$

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

$= \frac { \sum n ^ { 3 } + 2 \sum n ^ { 2 } + \sum n } { \sum n ^ { 3 } + \sum n ^ { 2 } }$

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

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

$= \frac { 3 n ^ { 2 } + 11 n + 10 } { 3 n ^ { 2 } + 7 n + 2 } = \frac { ( n + 2 ) ( 3 n + 5 ) } { ( n + 2 ) ( 3 n + 1 ) } = \frac { 3 n + 5 } { 3 n + 1 }$

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Question 24 Marks

Find the sum of the following series up to n terms: $\frac { 1 ^ { 3 } } { 1 } + \frac { 1 ^ { 3 } + 2 ^ { 3 } } { 1 + 3 } + \frac { 1 ^ { 3 } + 2 ^ { 3 } + 3 ^ { 3 } } { 1 + 3 + 5 } + \ldots \ldots$

Answer

Given: $\frac { 1 ^ { 3 } } { 1 } + \frac { 1 ^ { 3 } + 2 ^ { 3 } } { 1 + 3 } + \frac { 1 ^ { 3 } + 2 ^ { 3 } + 3 ^ { 3 } } { 1 + 3 + 5 } + \ldots \ldots$ up to n terms$\therefore a _ { n } = \frac { 1 ^ { 3 } + 2 ^ { 3 } + 3 ^ { 3 } + \ldots \ldots + n ^ { 3 } } { 1 + 3 + 5 + \ldots \ldots ( 2 n - 1 ) }$
$= {{\sum {{n^3}} } \over {{n \over 2}\left[ {2 + (n - 1)2} \right]}} = {{\sum {{n^3}} } \over {{n \over 2}(2n)}} = {{\sum {{n^3}} } \over {{n^2}}} = {{{n^2}{{(n + 1)}^2}} \over {4{n^2}}}$
$= \frac { 1 } { 4 } \left( n ^ { 2 } + 2 n + 1 \right)$
$\therefore {S_n} = \sum\limits_{k = 1}^n {{a_{_k}}} = \sum\limits_{k = 1}^n {{{{k^{^2}} + 2k + 1} \over 4}} $
$= \frac { 1 } { 4 }[(1^2 + 2.1 + 1) + (2^2 + 2.2 + 1) + (3^2 + 2.3 + 1) + ...... +(n^2 + 2n + 1)]$
$= {1 \over 4}\left[ {\sum {{n^2} + 2\sum {n + n} } } \right]$
$= \frac { 1 } { 4 } \left[ \frac { n ( n + 1 ) ( 2 n + 1 ) } { 6 } + \frac { 2 n ( n + 1 ) } { 2 } + n \right]$
$= \frac { n } { 4 } \left[ \frac { 2 n ^ { 2 } + 3 n + 1 + 6 n + 6 + 6 } { 6 } \right]$
$= \frac { n } { 24 } \left( 2 n ^ { 2 } + 9 n + 13 \right)$

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Question 34 Marks

If $S_1, S_2, S_3$ are the sum of first n natural no. their squares and their cubes respectively, show that $9 S _ { 2 } ^ { 2 } = S _ { 3 } \left( 1 + 8 S _ { 1 } \right)$.

Answer

$S _ { 1 } = \frac { n ( n + 1 ) } { 2 }$
$S _ { 2 } = \frac { n ( n + 1 ) ( 2 n + 1 ) } { 6 }$
$S _ { 3 } = \left( \frac { n ( n + 1 ) } { 2 } \right) ^ { 2 }$
R.H.S. = $S_3(1 + 8S_1)$
$= \frac { ( n ( n + 1 ) ) ^ { 2 } } { 2 } \left[ 1 + 8 \frac { n ( n + 1 ) } { 2 } \right]$
$= 9 \left( \frac { n ( n + 1 ) ( 2 n + 1 ) } { 6 } \right) ^ { 2 }$
$= 9 \mathrm { S } _ { 2 } ^ { 2 }$

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Question 44 Marks

If a, b, c are in A.P.; b, c, d are in G.P. and $\frac { 1 } { c } , \frac { 1 } { d } , \frac { 1 } { e }$ are in A.P., prove that a, c, e are in G.P.

Answer

Since, $a, b, c$ are in A.P.
$\therefore b-a=c-b$
$\Rightarrow 2 b=a+c$
$\Rightarrow b=\frac{a+c}{2}$
Since, b, c, $d$ are in G.P.
$\therefore \frac{c}{b}=\frac{d}{c}$
$\Rightarrow c^2=b d .$……….(i)
Also $\frac{1}{c}, \frac{1}{d}, \frac{1}{e}$ are in A.P.
$\therefore \frac{1}{d}-\frac{1}{c}=\frac{1}{e}-\frac{1}{d}$
$\Rightarrow \frac{2}{d}=\frac{1}{c}+\frac{1}{e}$
$\Rightarrow \frac{2}{d}=\frac{c+e}{c e}$
$\Rightarrow d=\frac{2 c e}{c+e}$
Putting values of $b$ and $d$ in eq. (i), $c^2=\left(\frac{c+a}{2}\right)\left(\frac{2 c e}{c+e}\right)$
$\Rightarrow c^2=\frac{c e(c+a)}{c+e}$
$\Rightarrow c^2(c+e)=e c(c+a)$
$\Rightarrow c^2+c e=c e+a e$
$\Rightarrow c^2=a e$ which shows that $a, c$, e are in G.P.

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Question 54 Marks

The ratio of the A.M. and G.M. of two positive numbers a and b is

Show that $a : b = \left( \begin{array} { c } { m + \sqrt { m ^ { 2 } - n ^ { 2 } } } \end{array} \right) : \left( m - \sqrt { m ^ { 2 } - n ^ { 2 } } \right)$

Answer

Given: $\frac { a + b } { 2 } : \sqrt { a b } = m : n$

$\Rightarrow \frac { a + b } { 2 \sqrt { a b } } = \frac { m } { n }$

By componendo and dividendo,

$\frac { a + b + 2 \sqrt { a b } } { a + b - 2 \sqrt { a b } } = \frac { m + n } { m - n }$

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

$\Rightarrow \frac { \sqrt { a } + \sqrt { b } } { \sqrt { a } - \sqrt { b } } = \frac { \sqrt { m + n } } { \sqrt { m - n } }$

Again by componendo and dividendo,

$\frac { \sqrt { a } + \sqrt { b } + \sqrt { a } - \sqrt { b } } { \sqrt { a } + \sqrt { b } - \sqrt { a } + \sqrt { b } } = \frac { \sqrt { m + n } + \sqrt { m - n } } { \sqrt { m + n } - \sqrt { m - n } }$

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

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

$\Rightarrow \frac { a } { b } = \frac { m + n + m - n + 2 \sqrt { ( m + n ) ( m - n ) } } { m + n + m - n - 2 \sqrt { ( m + n ) ( m - n ) } }$

$\Rightarrow \frac { a } { b } = \frac { 2 m + 2 \sqrt { ( m + n ) ( m - n ) } } { 2 m - 2 \sqrt { ( m + n ) ( m - n ) } }$

$\Rightarrow \frac { a } { b } = \frac { m + \sqrt { ( m + n ) ( m - n ) } } { m - \sqrt { ( m + n ) ( m - n ) } }$

Therefore, $a : b = \left( \begin{array} { c } { m + \sqrt { m ^ { 2 } - n ^ { 2 } } } \end{array} \right) : \left( m - \sqrt { m ^ { 2 } - n ^ { 2 } } \right)$

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