How do you simplify $n \sum k = 1 (3 k^{2} + 2 k)$ $\sum _ { k = 1} ^ { n } ( 3k ^ { 2} + 2k )$ ?

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Answer 1 · 2017-11-08T13:00:51

The answer is $= \frac{n}{2} (n + 1) (2 n + 3)$ $=n/2(n+1)(2n+3)$

Let's rewrite the summation

$n \sum k = 1 3 k^{2} + n \sum k = 1 2 k$ $sum_(k=1)^n3k^2+sum_(k=1)^n2k$

$= 3 n \sum k = 1 k^{2} + 2 n \sum k = 1 k$ $=3sum_(k=1)^nk^2+2sum_(k=1)^nk$

$= 3 \cdot \frac{n}{6} (n + 1) (2 n + 1) + 2 \cdot \frac{n}{2} (n + 1)$ $=3*n/6(n+1)(2n+1)+2*n/2(n+1)$

$= \frac{n}{2} (n + 1) (2 n + 1) + n (n + 1)$ $=n/2(n+1)(2n+1)+n(n+1)$

$= n (n + 1) (\frac{1}{2} (2 n + 1) + 1)$ $=n(n+1)(1/2(2n+1)+1)$

$= n (n + 1) (n + \frac{1}{2} + 1)$ $=n(n+1)(n+1/2+1)$

$= \frac{n}{2} (n + 1) (2 n + 3)$ $=n/2(n+1)(2n+3)$

How do you simplify n∑k=1(3k2+2k)\sum _ { k = 1} ^ { n } ( 3k ^ { 2} + 2k )?