If the sum of the first 10 terms of the series 11+14 4+ 21+24 4 + 31+34 4+ 41+44 4+ is mn, (m, n)=1, then m+n is equal to :
- 256
- 264
- 276
- 284
Answer
(3) 276
Solution
The general term is Tr = r1 + 4r4.
∵ 1 + 4r4 = (2r2 + 1)2 - (2r)2 = (2r2 - 2r + 1)(2r2 + 2r + 1) and (2r2 + 2r + 1) - (2r2 - 2r + 1) = 4r,
Tr = 14 [ 12r2 - 2r + 1 - 12r2 + 2r + 1 ]
Since 2(r + 1)2 - 2(r + 1) + 1 = 2r2 + 2r + 1, the terms cancel in pairs (telescoping), so
S10 = 14 [ 11 - 12(10)2 + 2(10) + 1 ] = 14 [1 - 1221 ] = 55221
∵ 221 = 13 17, (55, 221) = 1, so m = 55 and n = 221.
∴ m + n = 276
Answer: (3)