Question

The sum of the coefficients in the expansion of ${\left(1+x-3x^2\right)}^{3148}$ is

• A. $8$
• B. $7$
• C. $1$
• D. $-1$

Answer 1

The correct option is C

$1$

Explanation for the correct answer:

The expression ${\left(1+x-3x^2\right)}^{3148}$ can be expanded as

${\left(1+x-3x^2\right)}^{3148}=a_0+a_{1}x+a_2{x}^2+..........+a_{6296}{x}^{6296}$ $\left[\because \mathrm{highest}\mathrm{power}=x^{2^{3148}}\right]$

If we substitute $x=1$ in the above expansion we get the sum of the coefficients. Thus,

${\left(1+1-3\right)}^{3148}=a_0+a_1+a_2+............+a_{6296}$

$a_0+a_1+a_2+............+a_{6296}$ is the sum of all coefficients.

$\therefore$ sum of coefficients $={\left(-1\right)}^{3148}$

$=1$ (even power of $1$)

Hence, the sum of all coefficients in the expansion of ${\left(1+x-3x^2\right)}^{3148}$ is $1$

Hence, option C is the correct answer.