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 + . . . . . upto 10 t h term is

Step-by-step solution

S =   3 1 3 12 + 5 1 3 + 2 3 1 2 + 2 2 + 7 1 3 + 2 3 + 3 3 1 2 + 2 2 + 3 2 + … 10   t e r m s Here the general terms is T r = 2 r + 1 1 3 + 2 3 + … . + r 3 1 2 + 2 2 + … . + r 2 + 2 r + 1 r r + 1 2 2 r r + 1 2 r + 1 6 = 3 2   r ( r + 1 ) ⇒ S = ∑ r = 1 10 3 2 r r + 1 = 3 2 ∑ r = 1 10 r 2 + ∑ r = 1 10 r = 3 2 10.11.21 6 + 10.11 2 = 3 2 385 + 55 = 660