If the coefficient of $x$ in the expansion of $\left(ax^2+bx+c\right)(1-2x)^{26}$ is -56 and the coefficients of $x^2$ and $x^3$ are both zero, then $\mathrm{a}+\mathrm{b}+\mathrm{c}$ is equal to :

Question

Accepted Answer

B. 1. Expand $(ax^2+bx+c)(1-2x)^{26}$ term by term and extract the coefficient of $x$ from each piece.
Given the general term of $(1-2x)^{26}$ is $\binom{26}{k}(-2)^kx^k$, the coefficient of $x$ (from $bx$ times the constant term and $c$ times the $x^1$ term) is
$$
b-52c = -56
$$

2. Similarly extract the coefficient of $x^2$, which comes from $a$ times the constant term, $b$ times the $x^1$ term, and $c$ times the $x^2$ term of $(1-2x)^{26}$.
Given this coefficient is zero,
$$
a-52b+1300c = 0
$$

3. Similarly extract the coefficient of $x^3$, which involves the $x^0$, $x^1$, $x^2$ terms shifted appropriately.
Given this coefficient is also zero,
$$
-52a+1300b-20800c = 0 \quad\Rightarrow\quad -a+25b-400c=0
$$

4. Solve this system of three equations simultaneously for $a$, $b$, $c$.
Solving gives
$$
a=1300, \quad b=100, \quad c=3
$$

5. Sum these values.
Hence,
$$
a+b+c = 1300+100+3 = 1403
$$

Hence, $a+b+c=1403$, so the answer is option B.

If the coefficient of xxx in the expansion of (ax2+bx+c)(1−2x)26\left(ax^2+bx+c\right)(1-2x)^{26}(ax2+bx+c)(1−2x)26 is -56 and the coefficients of x2x^2x2 and x3x^3x3 are both zero, then a+b+c\mathrm{a}+\mathrm{b}+\mathrm{c}a+b+c is equal to :

If the coefficient of $x$ in the expansion of $\left(ax^2+bx+c\right)(1-2x)^{26}$ is -56 and the coefficients of $x^2$ and $x^3$ are both zero, then $\mathrm{a}+\mathrm{b}+\mathrm{c}$ is equal to :