Given, ∑_r=1^nr(r 1)(r+5) =∑_r=1^nr^3+4r^2 5r =∑ _r=1^nr^3+∑ _r=1^n4r^2 ∑ _r=1^n5r =14n^2(n+1)^2+23n(n+1)(2n+1) 52n(n+1) =n^2(n+1)^2· 3+ ...

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Byju's Answer

Standard XII

Mathematics

Sum of n Terms

Find ∑nr=1r...

Question

Find $\sum_{r = 1}^{n} r (r - 1) (r + 5)$

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Solution

Given,

$\sum_{r = 1}^{n} r (r - 1) (r + 5)$

$= \sum_{r = 1}^{n} r^{3} + 4 r^{2} - 5 r$

$= \sum_{r = 1}^{n} r^{3} + \sum_{r = 1}^{n} 4 r^{2} - \sum_{r = 1}^{n} 5 r$

$= \frac{1}{4} n^{2} {(n + 1)}^{2} + \frac{2}{3} n (n + 1) (2 n + 1) - \frac{5}{2} n (n + 1)$

$= \frac{n^{2} {(n + 1)}^{2} \cdot 3 + 8 n (n + 1) (2 n + 1) - 30 n (n + 1)}{12}$

$= \frac{3 n^{4} + 22 n^{3} - 3 n^{2} - 22 n}{12}$

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Find ∑nr=1r(r−1)(r+5)

Given,∑nr=1r(r−1)(r+5)=∑nr=1r3+4r2−5r=∑nr=1r3+∑nr=14r2−∑nr=15r=14n2(n+1)2+23n(n+1)(2n+1)−52n(n+1)=n2(n+1)2⋅3+8n(n+1)(2n+1)−30n(n+1)12=3n4+22n3−3n2−22n12

Find $\sum_{r = 1}^{n} r (r - 1) (r + 5)$