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JEE Main · 2023 · 15 Apr 2023 · Shift 1 · Binomial Theorem

Let

(a + b x + c x^{2})^{10} = \sum_{i = 0}^{20} p_{i} x^{i}, a, b, c \in N

. If

p_{1} = 20

and

p_{2} = 210

, then

2 (a + b + c)

is equal to

(A)8
(B)12✓ Correct
(C)15
(D)6

Answer: (B)

Solution

Coefficient of

x^{1}

:

p_{1} = 10 a^{9} b = 20

→

a^{9} b = 2

. Since

a, b \in N

,

a = 1, b = 2

.

Coefficient of

x^{2}

:

p_{2} = 10 a^{9} c + (2 10) a^{8} b^{2} = 210

.

Substitute

a = 1, b = 2

:

10 c + 45 \cdot 4 = 210

→

10 c = 30

→

c = 3

.

Thus

2 (a + b + c) = 2 (1 + 2 + 3) = 12

.

JEE Main · 2023 · 15 Apr 2023 · Shift 1 · Single correct option

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