(Each question 4 marks)

Question 14 Marks

Find the coefficient of $x^5$ in the product $(1 + 2x)^6 (1 -x)^7$ using binomial theorem.

Answer

Using binomial theorem
$(1+2 \mathrm{x})^6(1-\mathrm{x})^7=\left[{ }^6 C_0+{ }^6 C_1(2 x)+{ }^6 C_2(2 x)^2+{ }^6 C_3(2 x)^3+{ }^6 C_4(2 x)^4+{ }^6 C_5(2 x)^5+{ }^6 C_6(2 x)^6\right]$
${\left[{ }^7 C_0-{ }^7 C_1(x)+{ }^7 C_2(x)^2-{ }^7 C_3(x)^3+{ }^7 C_4(x)^4-{ }^7 C_5(x)^5+{ }^7 C_6(x)^6-{ }^7 C_7(x)^7\right]}$
$=\left[1+12 x+60 x^2+160 x^3+240 x^4+192 x^5+64 x^6\right]\left[1-7 x+21 x^2-35 x^3+35 x^4-21 x^5+7 x^6-x^7\right]$
$\therefore$ Coefficient of $x^5$ in the product
$=(1 \times-21)+(12 \times 35)+(60 \times-35)+(160 \times 21)+(240 \times-7)+(192 \times 1)$
$=-21+420-2100+3360-1680+192=171$

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Question 24 Marks

Find a, b and n in the expansion of $(a + b)^n$ if the first three terms of the expansion are $729, 7290$ and $30375$ respectively.

Answer

We have ${T_1}{ = ^n}{C_0}{a^n}{b^0} = 729$ . . . (i)
${T_2}{ = ^n}{C_1}{a^{n - 1}}b = 7290$ . . . (ii)
${T_3}{ = ^n}{C_2}{a^{n - 2}}{b^2} = 30375$ . . . (iii)
From (i) $a^n = 729$ . . . (iv)
From (ii) $na^{n-1} b = 7290$ . . . (v)
From (iii) $\frac{{n(n - 1)}}{2}{a^{n - 2}}{b^2}$ = 30375 . . . (vi)
Multiplying (iv) and (vi), we get
$\frac{{n(n - 1)}}{2}{a^{2n - 2}}{b^2} = 729 \times 30375$ . . . (vii)
Squaring both sides of (v) we get
$n^2a^{2n-2}b^2 = (7290)(7290)(viii)$
Dividing (vii) by (viii), we get
$\frac{{n(n - 1){a^{2n - 2}}{b^2}}}{{2{n^2}{a^{2n - 2}}{b^2}}} = \frac{{729 \times 30375}}{{7290 \times 7290}}$
$ \Rightarrow \frac{{(n - 1)}}{{2n}} = \frac{{30375}}{{72900}} \Rightarrow \frac{{n - 1}}{{2n}} = \frac{5}{{12}}$$ \Rightarrow 12n - 12 = 10n$
$ \Rightarrow 2n = 12$ $ \Rightarrow n = 6$
From (iv) ${a^6} = 729 \Rightarrow {a^6} = {(3)^6} \Rightarrow a = 3$
From (v) $6 \times {3^5} \times b = 7290 \Rightarrow b = 5$
Thus a = 3, b = 5 and n = 6.

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Question 34 Marks

Prove that the coefficient of $x^n$ in the expansion of $(1+x)^{2 n}$ is twice the coefficient of $x^n$ in the expansion of $(1+x)^{2 n-1}$.

Answer

We know that coefficient of $x^n$ in the expansion of $(1+x)^{2 n}$
$={ }^{2 n} C_n=\frac{(2 n)!}{n!n!}=\frac{2 n(2 n-1)!}{n(n-1)!n!}$
$=2 \frac{(2 n-1)!}{(n-1)!n!} \ldots \text { (i) }$
Also the coefficient of $x^n$ is the expansion of $(1+x)^{2 n-1}$
$={ }^{2 n-1} C_n=\frac{(2 n-1)!}{(n-1) n!} \ldots \text { (ii) }$
From (i) and (ii) we see that coefficient of $x^n$ in $(1+x)^{2 n}$ is twice the coefficient of $x^n$ in the expansion of $(1+x)^{2 n-1}$.

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Question 44 Marks

The coefficients of the $(r-1)^{\text {th }}, r^{\text {th }}$ and $(r+1)^{\text {th }}$ terms in the expansion of $(x+1)^n$ are in the ratio $1: 3: 5$. Find $n$ and $r$.

Answer

We know that coefficients of $(r-1)^{\text {th }}$, $r^{\text {th }}$ and $(r+1)^{\text {th }}$ terms in expansion of $(x+1)^n$ are ${ }^n C_{r-2},{ }^n C_{r-1}$ and ${ }^n C_r$ respectively.
$ \Rightarrow \frac{{^n{C_r}}}{{^n{C_{r - 1}}}} = \frac{5}{3}$ and $\frac{{^n{C_{r - 1}}}}{{^n{C_{r - 2}}}} = \frac{3}{1}$
$ \Rightarrow \frac{{n - r + 1}}{r} = \frac{5}{3}$ and $\frac{{n - r + 2}}{{r - 1}} = \frac{3}{1}\left[ {\because \frac{{^n{C_r}}}{{^n{C_{r - 1}}}} = \frac{{n - r + 1}}{r}} \right]$
$ \Rightarrow $ 3n - 8r + 3 = 0 and n - 4r + 5 = 0
Solving these for n and r, we get
n = 7 and r = 3

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MATHS (Each question 4 marks)

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