Binomial Theorem — JEE Main · 2023 · 06 Apr 2023 · Shift 2 Question with Solution

Among the statements:
(S1):

202 3^{2022} - 199 9^{2022}

is divisible by 8.
(S2):

13 (13)^{n} - 11 n - 13

is divisible by 144 for infinitely many

n \in N

Answer: (C)

Solution

For (S1): Write

202 3^{2022} = (1999 + 24)^{2022}

. By the binomial theorem,

(1999 + 24)^{2022} - 199 9^{2022}

equals

(1 2022) (1999)^{2021} (24) + (2 2022) (1999)^{2020} (24)^{2} + \dots

, and since

24 = 8 \times 3

, every term is divisible by 8. Hence (S1) is true.

For (S2): Expand

1 3^{n} = (1 + 12)^{n} = 1 + 12 n + (2 n) 1 2^{2} + \dots

. Then

13 (1 3^{n}) - 11 n - 13 = 13 (1 + 12 n + \dots) - 11 n - 13 = 145 n + (2 n) 1 2^{2} + \dots

. If we choose

n = 144 m

for any

m \in N

, then

145 n = 145 \cdot 144 m

is divisible by 144, and all higher binomial terms contain factors of

1 2^{2} = 144

. Thus the expression is divisible by 144 for infinitely many

n

. Hence (S2) is also true.

Both statements are correct.

JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Single correct option