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Joint Entrance Examination
Mathematics
Sequences and Series
Question 279
Question 279 of 519
Easy
NUMERICAL
The sum
∑
n
=
1
7
n
(
n
+
1
)
(
2
n
+
1
)
4
\displaystyle\sum_{n=1}^{7}\frac{n(n+1)(2n+1)}{4}
n
=
1
∑
7
4
n
(
n
+
1
)
(
2
n
+
1
)
is equal to \_\_\_\_\_\_\_\_. Answer: 504
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Official solution
\[\frac14\sum_{n=1}^{7}(2n^3+3n^2+n) = \frac14\left[2\left(\frac{7\times8}{2}\right)^2+3\cdot\frac{7\times8\times15}{6}+\frac{7\times8}{2}\right] = \frac{1568+420+28}{4} = 504.\]
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