If 1+(1+x)+(1+x)2+…..+(1+x)n=∑k=0n akxk, then

The given expression is 1+(1+x)+(1+x)2+…..+(1+x)n=∑k=0n akxkThe left hand side expression is sum of n+1 terms of GP with first term a=1 and the common ratio r=1+xhence, 1+(1+x)+(1+x)2+…..+(1+x)n=(1+x)n+1-1xsum of the coeffiecients is ∑k=0n ak=2n+1-1The coefficient of xn-2 is an-2=cn-1 n+1 =c2 n+1 =n+1n2The coefficients are like binomial coefficients, hence ap>ap−1 for p<n2Consider a92-a82a92-a82=a9-a8a9+a8 =C10 n+1+C9 n+1C10 n+1-C9 n+1 =C10 n+2C10 n+1-C9 n+1

If 1+(1+x)+(1+x)2+…..+(1+x)n=∑k=0n akxk, then

The given expression is 1+(1+x)+(1+x)2+…..+(1+x)n=∑k=0n akxkThe left hand side expression is sum of n+1 terms of GP with first term a=1 and the common ratio r=1+xhence, 1+(1+x)+(1+x)2+…..+(1+x)n=(1+x)n+1-1xsum of the coeffiecients is ∑k=0n ak=2n+1-1The coefficient of xn-2 is an-2=cn-1 n+1 =c2 n+1 =n+1n2The coefficients are like binomial coefficients, hence ap>ap−1 for p<n2Consider a92-a82a92-a82=a9-a8a9+a8 =C10 n+1+C9 n+1C10 n+1-C9 n+1 =C10 n+2C10 n+1-C9 n+1

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If $1 + (1 + x) + (1 + x)^{2} + \dots . . + (1 + x)^{n} = \sum_{k = 0}^{n} a_{k} x^{k}$ , then

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$\sum_{k = 0}^{n} a_{k} = 2^{n + 1}$

$a_{p} > a_{p - 1} for p < \frac{n}{2}$

${(a_{9})}^{2} - {(a_{8})}^{2} =^{n + 2} C_{10} (^{n + 1} C_{10} -^{n + 1} C_{9})$

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

answer is B, C, D.

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Detailed Solution

The given expression is $1 + (1 + x) + (1 + x)^{2} + \dots . . + (1 + x)^{n} = \sum_{k = 0}^{n} a_{k} x^{k}$

The left hand side expression is sum of $(n + 1)$ terms of GP with first term $a = 1$ and the common ratio $r = 1 + x$

hence,

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

sum of the coeffiecients is $\sum_{k = 0}^{n} a_{k} = 2^{n + 1} - 1$

The coefficient of $x^{n - 2}$ is

$\begin{array}{rcl} a_{n - 2} & = & {}^{n + 1}c_{n - 1} \\ = & {}^{n + 1}c_{2} \\ = & \frac{(n + 1) (n)}{2} \end{array}$

The coefficients are like binomial coefficients, hence $a_{p} > a_{p - 1} for p < \frac{n}{2}$

Consider ${(a_{9})}^{2} - {(a_{8})}^{2}$

$\begin{array}{rcl} {(a_{9})}^{2} - {(a_{8})}^{2} & = & (a_{9} - a_{8}) (a_{9} + a_{8}) \\ = & ({}^{n + 1}C_{10} + {}^{n + 1}C_{9}) ({}^{n + 1}C_{10} - {}^{n + 1}C_{9}) \\ = & ({}^{n + 2}C_{10}) ({}^{n + 1}C_{10} - {}^{n + 1}C_{9}) \end{array}$