The coefficient of $x^{8}$ in the expansion of
1 + (1 + x) + $(1 + x)^{2} + . . . . . . + (1 + x)^{n} (n \geq 8)$ is

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^{n + 1} C_{n - 8}

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^{n} C_{n - 8}

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Solution

The correct option is C $^{n + 1} C_{n - 8}$
Ans.(c)

$E =$ Sum of a G.P. of $(n + 1)$ terms

$1 [\frac{(1 + x)^{n + 1} - 1}{1 + x - 1}] = \frac{1}{x} [(1 + x)^{n + 1} - 1]$

Coefficient of $x^{8}$ in E is coefficient of $x^{9}$ in $(1 + x)^{n + 1}$ which is

$^{n + 1} C_{9} =^{n + 1} C_{(n + 1) - 9} =^{n + 1} C_{n - 8} \Rightarrow (c)$

as $^{n} C_{r} =^{n} C_{n - r}$

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The coefficient of x8 in the expansion of 1 + (1 + x) + (1+x)2+......+(1+x)n(n≥8) is

The correct option is C n+1Cn−8Ans.(c)E= Sum of a G.P. of (n+1) terms1[(1+x)n+1−11+x−1]=1x[(1+x)n+1−1]Coefficient of x8 in E is coefficient of x9 in (1+x)n+1 which isn+1C9=n+1C(n+1)−9=n+1Cn−8⇒(c)as nCr=nCn−r

The coefficient of $x^{8}$ in the expansion of
1 + (1 + x) + $(1 + x)^{2} + . . . . . . + (1 + x)^{n} (n \geq 8)$ is