JEE Main 2018 — Sequences And Series Question with Solution

Note :

Sum of square of first n odd terms

1² + 3² + 5² + . . . . .+ n² = $$\frac{n\left(2n-1\right)\left(2n+1\right)}{3}$$

Given,

1² + 2. 2² + 3² + 2.4² + 5² + 2.6² + . . . . . .

A = Sum of first 20 terms

$$\therefore$$A = 1² + 2.2² + 3² + 24² + 5² + 2.6² + . . . . . .20 terms

Arrange those terms this way,

A = [1² + 3² + 5² + . . . . . 10 terms] + [ 2.2² + 2.4² + 2.6² + . . . . 10 terms]

A = [ 1² + 3² + 5² + . . . . 10 terms ] + 2.² [ 1² + 2² + 3² + . . . .10 terms ]

A = $$\frac{10\times \left(2.10-1\right)\left(2.10+1\right)}{3}+2.2^2\left[\frac{10\times 11\times 21}{6}\right]$$

A = $$\frac{10\times 19\times 21}{3}+8\times \frac{10\times 11\times 21}{6}$$

A =70 $$\times$$ 19 + 70 $$\times$$ 44

A = 70 $$\times$$ 63

B = Sum of first 40 terms

Arrange those terms this way.

B = [1²+ 3² + 5² +. . . . 20 terms ] + [2.2² + 2.4² +. . . . . 20 terms ]

B = [1² + 3² + 5² + . . . . 20 terms] + 2.2² [1² + 2² + . . . 20 terms ]

B = $$\frac{20\times 39\times 41}{3}+8\times \frac{20\times 21\times 41}{6}$$

B = 260 $$\times$$ 41 + 560 $$\times$$ 41

B = 41 $$\times 820$$

$$\therefore$$ B $$-$$ 2A = 41 $$\times$$ 820 $$-$$ 2 $$\times$$ 70 $$\times$$ 63 = 24800

Given that B $$-$$ 2A = 100 $$\lambda$$

$$\therefore$$ 100 $$\lambda$$ = 24800

$$\Rightarrow \lambda$$ = 248