If (1+x)^{15}=a_0+a_1 x+ldots+a_{15} x^{15},t | vedclass

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

Question

(C) Given that $(1+x)^{15} = \sum_{r=0}^{15} {}^{15}C_r x^r = a_0 + a_1 x + \ldots + a_{15} x^{15}$.Comparing coefficients,we have $a_r = {}^{15}C_r$.

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(C) Given that $(1+x)^{15} = \sum_{r=0}^{15} {}^{15}C_r x^r = a_0 + a_1 x + \ldots + a_{15} x^{15}$.Comparing coefficients,we have $a_r = {}^{15}C_r$.We need to evaluate $\sum_{r=1}^{15} r \frac{a_r}{a_{r-1}}$.Using the property $\frac{{}^nC_r}{{}^nC_{r-1}} = \frac{n-r+1}{r}$,we get:$\frac{a_r}{a_{r-1}} = \frac{{}^{15}C_r}{{}^{15}C_{r-1}} = \frac{15-r+1}{r} = \frac{16-r}{r}$.Substituting this into the summation:$\sum_{r=1}^{15} r \left( \frac{16-r}{r} ight) = \sum_{r=1}^{15} (16-r)$.This is an arithmetic progression: $(16-1) + (16-2) + \ldots + (16-15) = 15 + 14 + \ldots + 1$.The sum of the first $n$ natural numbers is $\frac{n(n+1)}{2}$.For $n=15$,the sum is $\frac{15 	imes 16}{2} = 15 	imes 8 = 120$.

If $(1+x)^{15}=a_0+a_1 x+\ldots+a_{15} x^{15}$,then $\sum_{r=1}^{15} r \frac{a_r}{a_{r-1}}$ is equal to

Similar Questions

In the binomial expansion of $(a - b)^n, n \ge 5,$ the sum of the $5^{th}$ and $6^{th}$ terms is zero. Then $a/b$ equals:

If the coefficients of $x^{7}$ in $(x^{2}+\frac{1}{bx})^{11}$ and $x^{-7}$ in $(x-\frac{1}{bx^{2}})^{11}$,$b \neq 0$,are equal,then the value of $b$ is equal to:

The number of integral terms in the expansion of $(\sqrt{3} + \sqrt[8]{5})^{256}$ is

The coefficient of $x^{24}$ in the expansion of $(1+x^2)^{12}(1+x^{12})(1+x^{24})$ is

The sum of the coefficients of $x^{2/3}$ and $x^{-2/5}$ in the binomial expansion of $(x^{2/3} + \frac{1}{2}x^{-2/5})^9$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module