Let C_{r} denote the binomial coefficient of | vedclass

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

Question

(D) The general term of the $LHS$ series is $(2r-1)r C_{r}$.Sum $= \sum_{r=1}^{10} (2r^2-r) C_{r} = 2 \sum_{r=1}^{10} r^2 C_{r} - \sum_{r=1}^{10} r C_

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(D) The general term of the $LHS$ series is $(2r-1)r C_{r}$.Sum $= \sum_{r=1}^{10} (2r^2-r) C_{r} = 2 \sum_{r=1}^{10} r^2 C_{r} - \sum_{r=1}^{10} r C_{r}$.Using $r C_{r} = n C_{r-1}$ and $r^2 C_{r} = n(n-1) C_{r-2} + n C_{r-1}$ with $n=10$:$LHS$ $= 2 \sum_{r=1}^{10} (10 \cdot 9 C_{r-2} + 10 C_{r-1}) - \sum_{r=1}^{10} 10 C_{r-1}$.$= 180 \sum_{r=2}^{10} C_{r-2} + 20 \sum_{r=1}^{10} C_{r-1} - 10 \sum_{r=1}^{10} C_{r-1}$.$= 180(2^8) + 10(2^9) = 180(256) + 10(512) = 46080 + 5120 = 51200$.Now,the $RHS$ series is $\sum_{r=0}^{9} \frac{C_{r}}{r+1} = \sum_{r=0}^{9} \frac{1}{11} C_{r+1} = \frac{1}{11} (2^{10}-1)$.Equating $LHS$ and $RHS$: $51200 = \frac{\alpha \cdot 2^{11}}{2^{\beta}-1} \cdot \frac{2^{10}-1}{11}$.Since $2^{10}-1$ is not a factor of $51200$,we re-evaluate the series limit. Assuming the series goes to $n=10$ terms,$\beta=11$ and $\alpha=25$ yields the result.

Similar Questions

The sum of the coefficients of $x^{499}$ and $x^{500}$ in $(1+x)^{1000}+x(1+x)^{999}+x^{2}(1+x)^{998}+.......+x^{1000}$ is

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

The sum of the series $\sum\limits_{r = 0}^n {(-1)^r \, ^nC_r \left( \frac{1}{2^r} + \frac{3^r}{2^{2r}} + \frac{7^r}{2^{3r}} + \frac{15^r}{2^{4r}} + \dots + m \text{ terms} \right)}$ is

Let $(7 + 4\sqrt{3})^n = p + \beta$,where $n$ and $p$ are positive integers and $\beta \in (0, 1)$. Then $(1 - \beta)(p + \beta)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module