Binomial TheoremHard

Question

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 $a + b + c$ is equal to

Options

A.1300
B.1500
C.1403
D.1483

Solution

$\left( ax^{2} + bc + c \right)\sum_{r = 0}^{26}\mspace{2mu}\ ^{26}C_{r}( - 2x)^{r}$

Coeff. of $x^{2}:a.\ ^{26}C_{0}( - 2)^{0} + b.\ ^{26}C_{1}( - 2) +$ c. $\ ^{26}C_{2}( - 2)^{2} = 0$

$$\begin{array}{r} \Rightarrow a - 52b + 1300c = 0\#(1) \end{array}$$

Coeff. of $x^{3}:a.\ ^{26}C_{1}( - 2) + b.\ ^{26}C_{2}( - 2)^{2} +$ c. $\ ^{26}C_{3}( - 2)^{3} = 0$

$$\begin{array}{r} \Rightarrow - 52a + 1300b - 20800c = 0\#(2) \end{array}$$

Coeff. of $x = - 56$

$$\Rightarrow b.\ ^{26}C_{0}( - 2)^{0} + c.\ ^{26}C_{1}( - 2)^{1} = - 56$$

$$\begin{array}{r} b - 52c = - 56\#(3) \end{array}$$

After solving (1), (2) & (3)

$${a = 1300,\text{ }b = 100,c = 3 }{\Rightarrow a + b + c = 1403}$$

$$ $$

Create a free account to view solution

View Solution Free
Topic: Binomial Theorem·Practice all Binomial Theorem questions

More Binomial Theorem Questions