Question

Solve previous year Question of iit jee

Chapter :- sequence and series​

Answer 1

EXPLANATION.

Let A be the sum of first 20 terms.

B be the sum of first 40 terms.

Series = 1² + 2.2² + 3² + 2.4² + 5² + 2.6² + . . . . .

⇒ B - 2A = 100λ.

As we know that,

Let A be the sum of first 20 terms.

⇒ A = 1² + 2.2² + 3² + 2.4² + 5² + 2.6² + . . . . . .

⇒ A = (1² + 3² + 5² + . . . . . + 19²) + 2(2² + 4² + 6² + . . . . . + 20²).

⇒ A = (1² + 3² + 5² + . . . . . + 19²) + 2(2)(1² + 2² + 3² + . . . . . + 10²).

As we know that,

Sum of square of first n natural number = ∑r² = n(n + 1)(2n + 1)/6.

Using this formula in the equation, we get.

⇒ A = (20)(20 + 1)(2 x 20 + 1)/6 + 4(10)(10 + 1)(2 x 10 + 1)/6.

⇒ A = (20)(21)(41)/6 + 4 (10)(11)(21)/6.

⇒ A = 2870 + 1540.

⇒ A = 4410.

B be the sum of the first 40 terms.

⇒ B = (1² + 3² + 5² + . . . . . + 39²) + 2(2² + 4² + 6² + . . . . . + 40²).

⇒ B = (1² + 3² + 5² + . . . . . + 39²) + 4(1² + 2² + 3² + . . . . . + 20²).

⇒ B = (40)(40 + 1)(40 x 2 + 1)/6 + 4 (20)(20 + 1)(2 x 20 + 1)/6.

⇒ B = (40)(41)(81)/6 + 4 (20)(21)(41)/6.

⇒ B = 22140 + 11480.

⇒ B = 33620.

Given.

⇒ B - 2A = 100λ.

⇒ 33620 - 2(4410) = 100λ.

⇒ 33620 - 8820 = 100λ.

⇒ 24800 = 100λ.

⇒ λ = 248.

Option [B] is correct answer.