If the coefficient of x in the expansion of ( a x 2 + b x + c ) ( 1 ? 2 x ) 26 i s - 56 and the coefficients of x2 and x3 are both zero, then a+b+c is equal to : [2026]

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Solution

Q.
If the coefficient of x in the expansion of $(a x^{2} + b x + c) {(1 - 2 x)}^{26} is - 56$ and the coefficients of $x^{2} and 𝑥^{3}$ are both zero, then $a + b + c$ is equal to : [2026]

1 1300

2 1500

3 1403

4 1483

Ans.
(3)

$(a x^{2} + b x + c) \sum_{r = 0}^{26} {}^{26}{C_{r}} {(- 2 x)}^{r}$

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

$\Rightarrow a - 52 b + 1300 c = 0 . . . (1)$

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

$\Rightarrow - 52 a + 1300 b - 20800 c = 0 . . . (2)$

$Coeff. of x = - 56$

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

$\Rightarrow b - 52 c = - 56 . . . (3)$

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

$a = 1300, b = 100, c = 3$

$\Rightarrow a + b + c = 1403$

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