The sum of the series 1 1 − 3 ⋅ 1 2 + 1 4 + 2 1 − 3 ⋅ 2 2 + 2 4 + 3 1 − 3 ⋅ 3 2 + 3 4 + . ... up to 10 terms is

Step-by-step solution

General term of the given series is given by, T n = n 1 - 3 n 2 + n 4 ⇒ T n = n n 4 - 2 n 2 + 1 - n 2 ⇒ T n = n n 2 - 1 2 - n 2 ⇒ T n = n n 2 - 1 - n n 2 - 1 + n ⇒ T n = 1 2 n 2 - 1 + n - n 2 - 1 - n n 2 - 1 - n n 2 - 1 + n ⇒ T n = 1 2 1 n 2 - 1 - n - 1 n 2 - 1 + n ⇒ ∑ n = 1 10 T n = 1 2 1 - 1 - 1 1 + 1 1 - 1 5 + 1 5 - 1 11 + . . . + 1 89 - 1 109 ⇒ ∑ n = 1 10 T n = 1 2 - 1 - 1 109 ⇒ ∑ n = 1 10 T n = - 55 109