જો 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 $a_{39}=\frac{20 !(2)^{19} 	imes 1}{19 !}=20 	imes 2^{19}$

$\Rightarrow a_{1}+a_{3}+\ldots+a_{37}=2^{19}\left(2^{20}-1-20ight)$

$=2^{19}\left(2^{20}-21ight)$

જો $\left(1+x+2 x^{2}\right)^{20}=a_{0}+a_{1} x+a_{2} x^{2}+\ldots+a_{40} x^{40}$ હોય તો $a _{1}+ a _{3}+ a _{5}+\ldots+ a _{37}$ ની કિમંત મેળવો.

Similar Questions

$(2x + 1) (2x + 3) (2x + 5)----- (2x + 99)$ ના વિસ્તરણમાં $x^{49}$ નો સહગુણક મેળવો

જો $\frac{{ }^{11} C_1}{2}+\frac{{ }^{11} C_2}{3}+\ldots . .+\frac{{ }^{11} C_9}{10}=\frac{n}{m}$ જ્યાં ગુ. સા. અ. $\operatorname(n, m)=1$,હોય,તો $n+m$ .....................

$\sum_{\substack{i, j=0 \\ i \neq j}}^{n}{ }^{n} C_{i}{ }^{n} C_{j}$ ની કિમંત મેળવો.

ધારો કે $\alpha=\sum_{r=0}^n\left(4 r^2+2 r+1\right)^n C_r$ અને $\beta=\left(\sum_{r=0}^n \frac{{ }^n C_r}{r+1}\right)+\frac{1}{n+1} \cdot$ જો $140 < \frac{2 \alpha}{\beta}<281$ તો $n$ નું મૂલ્ય .......... છે.