Saved Bookmarks
| 1. |
hat has equal intercepts on the axes9. Find the sum of the first n term ofthe series 3+13+21+31+⌠|
|
Answer» Let us add 1 in the beginning for the series. We will subtract 1 from the sum of (N+1) terms to get the answer. Terms are : 1, 3, 7, 13, 21, 31, 43, 57, .. Differences between terms are : d_n = 2, 4, 6, 8, 12, 14 Sum of this series for n terms = 2 * n(n+1)/2 = n² + n So the series is : T1 = 1 T_n+1 = T1 + n² + n , for n >= 1 Rewrite it: T_n+1 = 1 + n + n² = (n+1)² - (n+1) + 1 => T_n = n² - n + 1 for n>= 1 So sum of the series for (n+1) terms is now: = Σ n² - Σ n + Σ 1 = (n+1) (n+2) (2 n + 3) / 6 - (n+1) (n+2)/2 + (n+1) = (n +1) / 6) [ 2 n² + 7n + 6 - 3 n - 6 + 6 ] = (n+1) (n² + 2 n + 3) / 3 = (n³ + 3 n² + 5 n + 3) / 3 Sum of the given series for n terms is now : (subtract 1) = (n³ + 3 n² + 5 n + 3 ) / 3 - 1 = (n³ + 3 n² + 5 n) / 3 |
|