The coefficients of $x^{r}$ in the expansion of
$1 + (1 + x) + (1 + x)^{2} + . . . . . + (1 + x)^{n}$ is $^{n + 1} C_{r + 1}$

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Solution

The given series is a G. P. of (n + 1) terms whose sum is
$\frac{(1 + x)^{n + 1} - 1}{1 + x - 1} = \frac{(1 + x)^{n + 1} - 1}{x}$
$∴$ coefficient of $x^{r} i n \frac{(1 + x)^{n + 1} - 1}{x}$ means coefficient of $x^{r + 1}$ in $[(1 + x)^{n + 1} - 1]$
It is clearly $^{n + 1} C_{r + 1}$

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Similar questions

Q. Assertion :The coefficient of

T_{r + 1}

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{(1 + x)}^{n + 1} =

sum of the coefficients of

T_{r}

T_{r + 1}

{(1 + x)}^{n}

Reason:

^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}

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The coefficients of xr in the expansion of 1+(1+x)+(1+x)2+.....+(1+x)n is n+1Cr+1

The given series is a G. P. of (n + 1) terms whose sum is (1+x)n+1−11+x−1=(1+x)n+1−1x∴ coefficient of xrin(1+x)n+1−1x means coefficient of xr+1 in [(1+x)n+1−1] It is clearly n+1Cr+1

The coefficients of $x^{r}$ in the expansion of
$1 + (1 + x) + (1 + x)^{2} + . . . . . + (1 + x)^{n}$ is $^{n + 1} C_{r + 1}$

The given series is a G. P. of (n + 1) terms whose sum is
$\frac{(1 + x)^{n + 1} - 1}{1 + x - 1} = \frac{(1 + x)^{n + 1} - 1}{x}$
$∴$ coefficient of $x^{r} i n \frac{(1 + x)^{n + 1} - 1}{x}$ means coefficient of $x^{r + 1}$ in $[(1 + x)^{n + 1} - 1]$
It is clearly $^{n + 1} C_{r + 1}$