Question

Assertion :Let $(1+x^2)=a_0+a_{1}x+a_2{x}^2+....+a_{36}{x}^{36}$
$a_0+a_3+a_6+...+a_{36}=\frac{2}{3}(2^{35}+1)$ Reason: $a_0+a_1+a_2+...+a_{36}=2^{36}$ and
$a_0+a_2+a_4+...+a_{36}=2^{35}$

• A. Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
• B. Statement-1 is True, Statement-2 is True; Statement-2 is Not a correct explanation for Statement-1
• C. Statement-1 is True, Statement-2 is False
• D. Statement-1 is False, Statement-2 is True

Answer 1

The correct option is B Statement-1 is True, Statement-2 is True; Statement-2 is Not a correct explanation for Statement-1

Let $x=1$

Therefore

$a_0+a_1+a_2...a_{36}=2^{36}$

Let $x=w$

Therefore

$a_0+a_{1}w+a_2{w}^2+a_3{w}^3+a_4{w}^4....a_{36}{w}^{36}=(1+w^2{)}^{36}=(-w{)}^{36}=1$

Let $x=w^2$

Therefore

$a_0+a_1{w}^2+a_2{w}^4+a_3{w}^6+a_4{w}^8....a_{36}{w}^{72}=(1+w^4{)}^{36}=(-w^2{)}^{36}=1$

Adding all them and then simplifying gives us

$3a_0+a_1(1+w+w^2)+a_2(1+w^2+w^4)+3a_3+....=2^{36}+2$

$3a_0+a_1(1+w+w^2)+a_2(1+w+w^2)+3a_3+....=2^{36}+2$

$3(a_0+a_3+a_6+a_9+...a_{36})=2(2^{35}+1)$

$a_0+a_3+a_6+a_9+...a_{36}=\frac{2\left(2^{35+1}\right)}{3}$