Question

2+7+14+......+ (n² + 2n - 1) =


​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

2+7+14+......+ (n² + 2n - 1) = n(2n²+9n+5)/6

Given,

nth term, Tn = n^2 + 2n - 2

To Find,

Sum of n terms

Solution,

The given sum of numbers is 2+7+14+...+ (n² + 2n - 1)

First term = 2 = 1² + 2 - 1

Therefore the required value, 2+7+14+...+ (n² + 2n - 1) is :

$\sum_{i=1}^{n} i^2 + 2i - 2\\\\\implies \sum_{i=1}^{n} i^2 + 2 \sum_{i=1}^{n} i - 2 \sum_{i=1}^{n} 1\\\\\implies \frac{n(n+1)(2n+1)}{6} + 2 \frac{n(n+1)}{2} - 2n\\\\\implies \frac{n(n+1)(2n+1)}{6} + n(n+1) - 2n$

$\\\implies \frac{(n^2 +n)(2n+1)}{6} + n^2 + n - 2n\\\\\implies \frac{2n^3 + n^2 + 2n^2 +n}{6} + n^2 - n\\\\\implies \frac{2n^3 + 3n^2 +n}{6} + n^2 - n\\\\\implies \frac{2n^3 + 3n^2 + n + 6(n^2-n)}{6}\\\\\implies \frac{2n^3 + 3n^2 + n + 6n^2 - 6n}{6}\\\\\implies \frac{2n^3 + 9n^2 + 5n}{6}$

#SPJ3