The sum- -n-17nn-12n-14 is equal to

∑_n=1^7n(n+1)(2n+1)4 =14∑_n=1^7(2n^3+3n^2+n) =14 [ 2∑_n=1^7n^3+3∑_n=1^7n^2+∑_n=1^7n ] =14 [ 2 × ( n(n+1)2 )^2+3×n(n+1)(2n+1)6+n(n+1)2 ] =14 [ 2 × ( 7× 82 ) ...

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Byju's Answer

Standard XII

Mathematics

Second Fundamental Theorem of Calculus

The sum, ∑n=1...

Question

The sum, $7 \sum n = 1 \frac{n (n + 1) (2 n + 1)}{4}$ is equal to

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Solution

$7 \sum n = 1 \frac{n (n + 1) (2 n + 1)}{4}$

$= \frac{1}{4} 7 \sum n = 1 (2 n^{3} + 3 n^{2} + n)$

$= \frac{1}{4} [2 7 \sum n = 1 n^{3} + 3 7 \sum n = 1 n^{2} + 7 \sum n = 1 n]$

$= \frac{1}{4} ⎡ ⎣ 2 \times {(\frac{n (n + 1)}{2})}^{2} + 3 \times \frac{n (n + 1) (2 n + 1)}{6} + \frac{n (n + 1)}{2} ⎤ ⎦$

$= \frac{1}{4} [2 \times {(\frac{7 \times 8}{2})}^{2} + 3 \times \frac{7 \times 8 \times 15}{6} + \frac{7 \times 8}{2}]$

$= \frac{1}{4} [2 \times 784 + 420 + 28] = 504$

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The sum, 7∑n=1n(n+1)(2n+1)4 is equal to

7∑n=1n(n+1)(2n+1)4 =147∑n=1(2n3+3n2+n) =14[27∑n=1n3+37∑n=1n2+7∑n=1n] =14⎡⎣2×(n(n+1)2)2+3×n(n+1)(2n+1)6+n(n+1)2⎤⎦ =14[2×(7×82)2+3×7×8×156+7×82] =14[2×784+420+28]=504

The sum, $7 \sum n = 1 \frac{n (n + 1) (2 n + 1)}{4}$ is equal to