Sum to n terms the series 1 + 5 + 14 + 30 + 55 + …

Given series is 1 + 5 + 14 + 30 + 55 + . . . Let S n be the sum of first n terms. Then, S n = 1 + 5 + 14 + 30 + 55 + … … + a n - 1 + a n (i) Rewrite the gives series with each term shifted by one place to the right ∴ S n = 1 + 5 + 14 + 30 + … + a n - 1 + a n (ii) equation (i) - equation (ii) gives, 0 = 1 + 4 + 9 + 16 + 25 + … n terms - a n ⇒ a n = 1 2 + 2 2 + 3 2 + 4 2 … … n terms We know that, Σ n 2 = n ( n + 1 ) ( 2 n + 1 ) 6 . ⇒ a n = n 6 2 n 2 + 3 n + 1 = 1 6 2 n 3 + 3 n 2 + n ∴ S n = ∑ k = 1 n a k = 1 3 ∑ k = 1 n k 3 + 1 2 ∑ k = 1 n k 2 + 1 6 ∑ k = 1 n k ⇒ 1 3 n 2 ( n + 1 ) 2 4 + 1 2 × n ( n + 1 ) ( 2 n + 1 ) 6 + 1 6 n ( n + 1 ) 2 ⇒ n ( n + 1 ) 12 [ n ( n + 1 ) + 2 n + 1 + 1 ] ⇒ n ( n + 1 ) 12 n 2 + 3 n + 2 ⇒ n ( n + 1 ) ( n + 1 ) ( n + 2 ) 12 ⇒ 1 12 n ( n + 1 ) 2 ( n + 2 ) ∴ Sum of n terms of the series is 1 12 n ( n + 1 ) 2 ( n + 2 ) .

Sum to n terms the series whose n t h term is 1 4 n 2 - 1

Let a n be the n t h terms of given series. ⇒ a n = 1 4 n 2 - 1 = 1 ( 2 n - 1 ) ( 2 n + 1 ) = 1 2 1 2 n - 1 - 1 2 n + 1 → ( 1 ) putting n = 1 , 2 , 3 , … , n in equation ( 1 ) , we have ⇒ a 1 = 1 2 1 1 - 1 3 , a 2 = 1 2 1 3 - 1 5 , . . . . . . . , a n = 1 2 1 2 n - 1 - 1 2 n + 1 On adding column wise, we have a 1 + a 2 … … + a n = S n ⇒ 1 2 1 - 1 2 n + 1 ⇒ 1 2 2 n + 1 - 1 2 n + 1 = n 2 n + 1 ∴ Sum of n terms of the given series is n 2 n + 1 .

Find the n t h term and the sum of n terms of the series 1 1 + 1 1 + 2 + 1 1 + 2 + 3 + …

Let a n be the n t h term and S n be the sum of first n terms of given series ⇒ a n = 1 1 + 2 + 3 … … + n ⇒ 1 Σ n = 1 ∑ k = 1 n k = 2 n ( n + 1 ) ∴ a n = 2 1 n - 1 n + 1 (i) putting n = 1 , 2 , 3 , … , n in equation (i), we have a 1 = 2 1 - 1 2 , a 2 = 2 1 2 - 1 3 , . . . . . . . , a n = 2 1 n - 1 n + 1 On adding column wise, we have a 1 + a 2 + … … + a n = S n ⇒ 2 1 - 1 n + 1 = 2 n n + 1 ∴ Sum of n terms of the given series is 2 n n + 1 .