Evaluate- -r-1n 3r-1r-1

∑_r=1^n(3r 1)(r+1) =∑_r=1^n(3r^2+2r 1) =∑_r=1^n3r^2+∑_r=1^n2r ∑_r=1^n1 =3∑_r=1^nr^2+2∑_r=1^nr 1∑_r=1^n1 =3 ( n(n+1)(2n+1)6 )+2 ( n(n+1) ...

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Byju's Answer

Standard XII

Mathematics

Summation by Sigma Method

Evaluate: ∑...

Question

Evaluate: $n \sum r = 1 (3 r - 1) (r + 1)$

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Solution

$\sum_{r = 1}^{n} (3 r - 1) (r + 1)$

$= \sum_{r = 1}^{n} (3 r^{2} + 2 r - 1)$

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

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

$= 3 (\frac{n (n + 1) (2 n + 1)}{6}) + 2 (\frac{n (n + 1)}{2}) - n$

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

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

$= \frac{2 n^{3} + n^{2} + 2 n^{2} + n + 2 n^{2}}{2}$

$= \frac{2 n^{3} + 5 n^{2} + n}{2}$

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Evaluate: n∑r=1(3r−1)(r+1)

∑nr=1(3r−1)(r+1)=∑nr=1(3r2+2r−1)=∑nr=13r2+∑nr=12r−∑nr=11=3∑nr=1r2+2∑nr=1r−1∑nr=11=3(n(n+1)(2n+1)6)+2(n(n+1)2)−n=n(n+1)(2n+1)2+n(n+1)−n=(n2+n)(2n+1)2+n2=2n3+n2+2n2+n+2n22=2n3+5n2+n2

Evaluate: $n \sum r = 1 (3 r - 1) (r + 1)$