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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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

Similar Questions

The coefficient of $x$ in the expansion of $[\sqrt{1 + x^2} - x]^{-1}$ in ascending powers of $x$,when $|x| < 1$,is

If $r, k, p \in W,$ then $\sum\limits_{r + k + p = 10} {{}^{30}{C_r} \cdot {}^{20}{C_k} \cdot {}^{10}{C_p}} $ is equal to -

If the sum of the coefficients in the expansion of $(\alpha x^2 - 2x + 1)^{35}$ is equal to the sum of the coefficients in the expansion of $(x - \alpha y)^{35}$,then $\alpha = $

If $C_{j}$ stands for ${ }^{n} C_{j}$,then $\frac{C_0}{2} + \frac{C_1}{2 \cdot 2^2} + \frac{C_2}{3 \cdot 2^3} + \ldots + \frac{C_{n}}{(n+1) 2^{n+1}} = $

If $\sum_{r=0}^{10} \left( \frac{10^{r+1}-1}{10^r} \right) \cdot {}^{11}C_{r+1} = \frac{\alpha^{11}-11^{11}}{10^{10}}$,then $\alpha$ is equal to :

Vedclass Test Series

Exam Paper Generator

Online Exam Module