Let { }^{n} C_{r} denote the binomial coeffic | vedclass

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

Question

(A) Given the sum $\sum_{k=0}^{10}(4+3k){ }^{10} C _{ k } = \alpha \cdot 3^{10} + \beta \cdot 2^{10}.$We know that $\sum_{k=0}^{n} {}^{n}C_k = 2^n$ an

Vedclass · Accepted Answer

$19$

Vedclass · Answer

(A) Given the sum $\sum_{k=0}^{10}(4+3k){ }^{10} C _{ k } = \alpha \cdot 3^{10} + \beta \cdot 2^{10}.$We know that $\sum_{k=0}^{n} {}^{n}C_k = 2^n$ and $\sum_{k=0}^{n} k \cdot {}^{n}C_k = n \cdot 2^{n-1}.$Expanding the sum: $\sum_{k=0}^{10} 4 \cdot {}^{10}C_k + 3 \sum_{k=0}^{10} k \cdot {}^{10}C_k.$$= 4 \cdot 2^{10} + 3 \cdot (10 \cdot 2^{9}).$$= 4 \cdot 2^{10} + 30 \cdot 2^{9} = 4 \cdot 2^{10} + 15 \cdot 2^{10} = 19 \cdot 2^{10}.$Comparing with $\alpha \cdot 3^{10} + \beta \cdot 2^{10},$ we get $\alpha = 0$ and $\beta = 19.$Therefore,$\alpha + \beta = 0 + 19 = 19.$

Let ${ }^{n} C_{r}$ denote the binomial coefficient of $x^{r}$ in the expansion of $(1+ x )^{ n }.$ If $\sum_{ k =0}^{10}\left(2^{2}+3 k \right){ }^{10} C _{ k }=\alpha \cdot 3^{10}+\beta \cdot 2^{10},$ where $\alpha, \beta \in R,$ then $\alpha+\beta$ is equal to ....... .

Similar Questions

The ratio of the coefficients of the terms $x^{n-r}a^r$ and $x^ra^{n-r}$ in the binomial expansion of $(x+a)^n$ is:

If ${}^{21}C_1 + 3 \cdot {}^{21}C_3 + 5 \cdot {}^{21}C_5 + \dots + 19 \cdot {}^{21}C_{19} + 21 \cdot {}^{21}C_{21} = k$,then the number of prime factors of $k$ is:

The sum of the coefficients of the last $19$ terms in the binomial expansion of $(1+x)^{37}$ is

If $\sum_{r=0}^5 \frac{{}^{11}C_{2r+1}}{2r+2} = \frac{m}{n}$,$\text{gcd}(m, n) = 1$,then $m - n$ is equal to . . . . . .

The sum of the series $\binom{20}{0} - \binom{20}{1} + \binom{20}{2} - \binom{20}{3} + \dots + \binom{20}{10}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module