Exams › JEE Main › Maths

Let T0, T1, T2,..., Tn denote the successive terms in the expansion of (x + a)ⁿ. Then the value of (T0 - T2 + T4 -...)² + (T1 - T3 + T5 -...)² is:

1. (x² + a²)²
2. (x² + a²)ⁿ
3. (x² + a²)^(1/n)
4. (x² + a²)^(-1/n)

Correct answer: (x² + a²)ⁿ

Solution

(x+ia)^n has real part T0-T2+T4-... and imaginary part T1-T3+T5-.... So the sum of their squares is |x+ia|^(2n) = (x^2+a^2)^n.

Related JEE Main Maths questions

• Let Pn = cosⁿ θ + sinⁿ θ. If Pn - Pn-2 = k Pn-4, then the value of k is:
• For any natural number n, the expression x^(2n - 1) + y^(2n - 1) is divisible by which of the following?
• For a natural number n, which of the following always divides the expression x^(n+1) + (x + 1)^(2n−1)?
• For a positive integer n, the expression 5^(2n+2) − 24n − 25 is divisible by which of the following?
• For every natural number n, the expression 41ⁿ − 14ⁿ is divisible by which of the following?
• Let C0, C1, C2,..., C15 denote the binomial coefficients in the expansion of (1 + x)¹⁵. Then the value of C1/C0 + C2/C1 + C3/C2 +... + C15/C14 is:

⚔️ Practice JEE Main Maths free + battle 1v1 →