If (1 + x - 2x^2)^6 = 1 + a_1x + a_2x^2 + ... | vedclass

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

Question

(B) Given the expansion: $(1 + x - 2x^2)^6 = 1 + a_1x + a_2x^2 + .... + a_{12}x^{12}$.Let $f(x) = (1 + x - 2x^2)^6 = 1 + a_1x + a_2x^2 + a_3x^3 + ....

Vedclass · Accepted Answer

$31$

Vedclass · Answer

(B) Given the expansion: $(1 + x - 2x^2)^6 = 1 + a_1x + a_2x^2 + .... + a_{12}x^{12}$.Let $f(x) = (1 + x - 2x^2)^6 = 1 + a_1x + a_2x^2 + a_3x^3 + .... + a_{12}x^{12}$.For $x = 1$: $f(1) = (1 + 1 - 2)^6 = 0^6 = 0 = 1 + a_1 + a_2 + a_3 + .... + a_{12}$.For $x = -1$: $f(-1) = (1 - 1 - 2(-1)^2)^6 = (-2)^6 = 64 = 1 - a_1 + a_2 - a_3 + .... + a_{12}$.Adding the two equations:$f(1) + f(-1) = 0 + 64 = 2(1 + a_2 + a_4 + a_6 + a_8 + a_{10} + a_{12})$.$64 = 2(1 + a_2 + a_4 + a_6 + a_8 + a_{10} + a_{12})$.$32 = 1 + a_2 + a_4 + a_6 + a_8 + a_{10} + a_{12}$.Therefore,$a_2 + a_4 + a_6 + a_8 + a_{10} + a_{12} = 32 - 1 = 31$.

If $(1 + x - 2x^2)^6 = 1 + a_1x + a_2x^2 + .... + a_{12}x^{12}$,then the expression $a_2 + a_4 + a_6 + .... + a_{12}$ has the value

Similar Questions

If $f(x) = x^n$,then the value of $f(1) - \frac{f'(1)}{1!} + \frac{f''(1)}{2!} - \frac{f'''(1)}{3!} + ...... + \frac{(-1)^n f^n(1)}{n!}$ is

The sum of the coefficients of the terms of degree $m$ in the expansion of $(1 + x)^n(1 + y)^n(1 + z)^n$ is:

The coefficient of $x^{19}$ in the polynomial $(x-1)(x-2^1)(x-2^2) \dots (x-2^{19})$ is:

If $\alpha$ and $\beta$ are the coefficients of $x^{4}$ and $x^{2}$ respectively in the expansion of $(x+\sqrt{x^{2}-1})^{6}+(x-\sqrt{x^{2}-1})^{6}$,then:

If the number of terms in the binomial expansion of $(2x + 3)^{3n}$ is $22$,then the value of $n$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module