The sum 3 × 1 3 1 2 + 5 × ( 1 3 + 2 3 ) 1 2 + 2 2 + 7 × ( 1 3 + 2 3 + 3 3 ) 1 2 + 2 2 + 3 2 +…

The sum 3×1312+5×(13+23)12+22+7×(13+23+33)12+22+32+..... upto 10th term is
  1. 660
  2. 600
  3. 620
  4. 680

Solution

S= 31312+513+2312+22+713+23+3312+22+32+10 terms
Here the general terms is
Tr=2r+113+23+.+r312+22+.+r2+2r+1rr+122rr+12r+16
=32 r(r+1)
S=r=11032rr+1=32r=110r2+r=110r
=3210.11.216+10.112=32385+55=660

Asked in: JEE Main 2019 (10 Apr Shift 1)

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