Question

If ${(1+x+x^2)}^{20}=\sum _{r=0}^{40}{a}_r{x}^r$ ,then the value of$a_1+a_3+a_5+\cdots a_{37}$ equals

• A. $3^{20}(3^{19}-21)$
• B. $\frac{3^{20}-41}{2}$
• C. $2^{20}(2^{19}-21)$
• D. $3^{19}(3^{20}-21)$

Answer 1

The correct option is B $\frac{3^{20}-41}{2}$

Given $(1+x+x^2{)}^{20}={\sum }_{r=0}^{40}{a}_r{x}^r$..........................(1)

Let $S=a_1+a_3+a_5+......a_{37}=\frac{1}{2}\left[\right(a_0+a_1+....a_{40})+(a_0-a_1+......+a_{40}\left)\right]-a_{39}$

Substitute x=1 and x=-1 in eqn (1)

$(1+1+1^2{)}^{20}={\sum }_{r=0}^{40}{a}_r$

$(1-1+(-1{)}^2{)}^{20}={\sum }_{r=0}^{40}(-1{)}^r{a}_r$

$S=\frac{3^{20}-1}{2}-a_{39}$

now we need to find $a_{39}$

Since the given expression is symmetrical $\Rightarrow a_1=a_{39}$

$a_1$ can be obtained by differentiating the eqn (1) and substituting $x=0$.

$a_1=\frac{d\left(\right(1+x+x^2{)}^{20})}{dx}=20(1+x+x^2{)}^{20-1}\frac{d(1+x+x^2)}{dx}$ at x=0 ($\because \frac{d\left(x^n\right)}{dx}=n{x}^{n-1}$)

$a_1=20(1+0+0{)}^{19}(1+2.0)=20$.

$S=\frac{3^{20}-1}{2}-20=\frac{3^{20}-41}{2}$