Let left(1+x+2 x^{2}right)^{20}=a_{0}+a_{1} x | vedclass

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

Question

$\left(1+x+2 x^{2}\right)^{20}=a_{0}+a_{1} x+\ldots .+a_{40} x^{40}$ put $x=$ $1,-1$

$\Rightarrow a_{0}+a_{1}+a_{2}+\ldots .+a_{40}=2^{20}$

$a_{

Vedclass · Accepted Answer

$2^{19}\left(2^{20}-21\right)$

Vedclass · Answer

$\left(1+x+2 x^{2}ight)^{20}=a_{0}+a_{1} x+\ldots .+a_{40} x^{40}$ put $x=$ $1,-1$

$\Rightarrow a_{0}+a_{1}+a_{2}+\ldots .+a_{40}=2^{20}$

$a_{0}-a_{1}+a_{2}+\ldots+a_{40}=2^{20}$

$\Rightarrow a_{1}+a_{3}+\ldots+a_{39}=\frac{4^{20}-2^{20}}{2}$

$\Rightarrow a_{1}+a_{3}+\ldots+a_{37}=2^{39}-2^{19}-a_{39}$

here $a_{39}=\frac{20 !(2)^{19} 	imes 1}{19 !}=20 	imes 2^{19}$

$\Rightarrow a_{1}+a_{3}+\ldots+a_{37}=2^{19}\left(2^{20}-1-20ight)$

$=2^{19}\left(2^{20}-21ight)$

Let $\left(1+x+2 x^{2}\right)^{20}=a_{0}+a_{1} x+a_{2} x^{2}+\ldots+a_{40} x^{40}$ then $a _{1}+ a _{3}+ a _{5}+\ldots+ a _{37}$ is equal to

Similar Questions

If n is a positive integer and ${C_k} = {\,^n}{C_k}$, then the value of ${\sum\limits_{k = 1}^n {{k^3}\left( {\frac{{{C_k}}}{{{C_{k - 1}}}}} \right)} ^2}$ =

$\frac{1}{{1!(n - 1)\,!}} + \frac{1}{{3!(n - 3)!}} + \frac{1}{{5!(n - 5)!}} + .... = $

Let $a =$ Minimum $\{x^2 + 2x + 3, x \in R\}$ and $b = \mathop {\lim }\limits_{\theta \to 0} \frac{{1 - \cos \theta }}{{{\theta ^2}}}$ The value of $\sum\limits_{r = 0}^n {{a^r}.{b^{n - r}}} $ is

The sum of coefficients of integral power of $x$ in the binomial expansion ${\left( {1 - 2\sqrt x } \right)^{50}}$ is :

What is the sum of the coefficients of ${({x^2} - x - 1)^{99}}$