Let a 1 = b 1 = 1 and a n = a n - 1 + ( n - 1 ) , b n = b n - 1 + a n - 1 , ∀ n ≥ 2 . If S = ∑ n = 1 10 b n 2 n and T = ∑ n = 1 8 n 2 n - 1 then 2 7 ( 2 S - T ) is equal to

Step-by-step solution

Given: a n = a n - 1 + n - 1 b n = b n - 1 +   a n - 1 And, S = b 1 2 + b 2 2 2 + … … . . + b 9 2 9 + b 10 2 10     . . . 1 ⇒ S 2 =    b 1 2 2 + b 2 2 3 + … … . . + b 9 2 10 + b 10 2 11       . . . 2 Subtracting 1 and 2 , we get S 2 = b 2 + b 2 - b 1 2 2 + b 3 - b 2 2 3 + . . . . + b 10 - b 9 2 10 - b 10 2 11 ⇒ S 2 = b 1 2 + a 1 2 2 + a 2 2 3 … … . + a 9 2 10 - b 10 2 11 ⇒ S = b 1 - b 10 2 10 + a 1 2 + a 2 2 2 … &#82