The sum of the series aC_0 + (a + b)C_1 + (a | vedclass

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

Question

Simplifying we get

$a\left[{ }^{n} C_{0}+{ }^{n} C_{1}+{ }^{n} C_{2}+\ldots .^{n} C_{n}\right]+b\left[{ }^{n} C_{1}+2{ }^{n} C_{2}+3{ }^{n} C_{3}+\

Vedclass · Accepted Answer

$(2a + nb)2^{n - 1}$

Vedclass · Answer

Simplifying we get

$a\left[{ }^{n} C_{0}+{ }^{n} C_{1}+{ }^{n} C_{2}+\ldots .^{n} C_{n}\right]+b\left[{ }^{n} C_{1}+2{ }^{n} C_{2}+3{ }^{n} C_{3}+\ldots n^{n} C_{n}\right]$

$=a 2^{n}+b\left(\frac{d(1+x)^{n}}{d x}\right)_{x=1}$

$=a 2^{n}+b\left(n(1+x)^{n-1}\right)_{x=1}$

$=a 2^{n}+b\left(n 2^{n-1}\right)$

$=(2 a+b n) 2^{n-1}$

The sum of the series $aC_0 + (a + b)C_1 + (a + 2b)C_2 + ..... + (a + nb)C_n$ is where $Cr's$ denotes combinatorial coefficient in the expansion of $(1 + x)^n, n \in N$

Similar Questions

Let $(1+2 x)^{20}=a_0+a_1 x+a_2 x^2+\ldots+a_{20} x^{20}$.Then $3 a_0+2 a_1+3 a_2+2 a_3+3 a_4+2 a_5+\ldots+2 a_{19}+3 a_{20}$ equals

If $C_r= ^{100}{C_r}$ , then $1.C^2_0 - 2.C^2_1 + 3.C^2_3 - 4.C^2_0 + 5.C^2_4 - .... + 101.C^2_{100}$ is equal to

Let $(1 + x)(1 + x + x^2)(1 + x + x^2 + x^3)\,\, ......\,\,$$(1 + x + x^2 + ..... + x^{30}) = $$a_0 + a_1x + a_2x^2$ .....$+$ $a_{465}x^{465}$, then sum of $a_0 + a_2 + a_4 + ......... +$ is

If $A$ denotes the sum of all the coefficients in the expansion of $\left(1-3 x+10 x^2\right)^n$ and $B$ denotes the sum of all the coefficients in the expansion of $\left(1+x^2\right)^n$, then :

Let $\left(2 x ^{2}+3 x +4\right)^{10}=\sum \limits_{ r =0}^{20} a _{ r } x ^{ r } \cdot$ Then $\frac{ a _{7}}{ a _{13}}$ is equal to