In the expansion of the expression 1 + (1 + x | vedclass

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(A) The given expression is a geometric progression $(G.P.)$ with $n+1$ terms,where the first term $a = 1$ and the common ratio $r = (1 + x)$.The sum

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$^{n + 1}C_{k + 1}$

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(A) The given expression is a geometric progression $(G.P.)$ with $n+1$ terms,where the first term $a = 1$ and the common ratio $r = (1 + x)$.The sum of the $G.P.$ is given by $S = \frac{a(r^{n+1} - 1)}{r - 1}$.Substituting the values,we get $E = \frac{1((1 + x)^{n + 1} - 1)}{(1 + x) - 1} = \frac{(1 + x)^{n + 1} - 1}{x}$.This simplifies to $E = x^{-1} \{(1 + x)^{n + 1} - 1\}$.To find the coefficient of $x^k$ in $E$,we need to find the coefficient of $x^{k+1}$ in the expansion of $(1 + x)^{n + 1} - 1$.Using the Binomial Theorem,the general term in $(1 + x)^{n + 1}$ is $^{n+1}C_r x^r$.Thus,the coefficient of $x^{k+1}$ is $^{n+1}C_{k+1}$.Therefore,the coefficient of $x^k$ in the expression is $^{n+1}C_{k+1}$.

In the expansion of the expression $1 + (1 + x) + (1 + x)^2 + \dots + (1 + x)^n$,the coefficient of $x^k$ $(0 \le k \le n)$ is

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