JEE Main | 2018Progression (Sequence and Series)Hard

Question

Let A be the sum of the first 20 terms and B be the sum of the first 40 terms of the series 12 + 2·22 + 32 + 2·42 + 52 + 2·62 + .......
If B - 2A = 100λ, then λ is equal to :

Options

A.

248

B.

464

C.

496

D.

232

Solution

B - 2A = ∑r=140Tr-2∑r=120Tr
= ∑r=2140Tr-∑r=120Tr
B - 2A = (212 + 2.222 + 232 + 2.242 + ...... + 402)
- (12 + 2.22 + 32 + 2.42 ..... + 202)
= 20[22 + 2.24 + 26 + 2.28 + ........ + 60]
= 20 22+24+26......60⏟20 terms+24+28+......+60⏟10 terms
= 20 202(22+60)+102(24+60)
= 10[20.82 + 10.84]
= 100[164 + 84] = 100.248

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