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The sum, \sum_{n=1}^{7}\frac{n(n+1)(2n+1)}{4} is equal to  _______.
Option: 1 1024  
Option: 3 810
Option: 5 504
Option: 7 1008
 

Answers (1)

best_answer

Sum of Common Series

Sum of the square of first n-natural numbers

            \\\mathrm{1^2+2^2+3^2+4^2+...........+n^2=\frac{n(n+1)(2n+1)}{6}}Sum of the cube of first n-natural numbers

           \\\mathrm{1^3+2^3+3^3+4^3+...........+n^3=\left \{ \frac{n(n+1)}{2} \right \}^2}

-

Now,

\begin{aligned} &\frac{1}{4}\left[\sum_{n=1}^{7}\left(2 n^{3}+3 n^{2}+n\right)\right]\\ &\frac{1}{4}\left[2\left(\frac{7.8}{2}\right)^{2}+3\left(\frac{7.8 .15}{6}\right)+\frac{7.8}{2}\right]\\ &\frac{1}{4}[2 \times 49 \times 16+28 \times 15+28]\\ &\frac{1}{4}[1568+420+28]=504 \end{aligned}

Correct Option (3)

Posted by

vishal kumar

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