Question

Let ${(1+x)}^{36}=a_0+a_{1}x+a_2{x}^2+...a_{36}{x}^{36}$. Then
$a_0+a_3+a_6+....+a_{36}$ is equal to

• A. $\frac{2}{3}(2^{35}+1)$
• B. $2^{35}$
• C. $2(2^{35}+1)$
• D. None of these

Answer 1

The correct option is A $\frac{2}{3}(2^{35}+1)$

We have
$a_0+a_1+a_2{x}^2+....+a_{36}{x}^{36}={(1+x)}^{36}...\left(i\right)$

Putting $x=1.\omega ,{\omega }^2$ successively, we get

$(a_0+a_3+a_6+....+a_{36})+(a_1+a_4+a_7+.....+a_{34})+(a_2+a_5+.....+a_{35})=2^{36}......\left(ii\right)$
$(a_0+a_3+a_6+....+a_{36})+\omega (a_1+a_4+a_7+.....+a_{34})+{\omega }^2(a_2+a_5+.....+a_{35})=\mathrm{1......}\left(iii\right)$ $[\because {\omega }^3=1]$
$(a_0+a_3+a_6+....+a_{36})+{\omega }^2(a_1+a_4+a_7+.....+a_{34})+\omega (a_2+a_5+.....+a_{35})=\mathrm{1.....}\left(iv\right)$

Adding Eqs. (i), (ii), (iv) we get

$3(a_0+a_3+a_6+....+a_{36})=2(2^{35}+1)$ $[1+\omega +{\omega }^2=0]$

$\Rightarrow a_0+a_3+a_6+....+a_{36}=\frac{2}{3}(2^{35}+1)$