Sum of Summations

Algebra Level 3

$\sum _{ n=1 }^{ 12 }{ { n }^{ 2 } } +\sum _{ n=2 }^{ 12 }{ { n }^{ 2 } } +\sum _{ n=3 }^{ 12 }{ { n }^{ 2 } } +\cdots+\sum _{ n=11 }^{ 12 }{ { n }^{ 2 } } +\sum _{ n=12 }^{ 12 }{ { n }^{ 2 } } = \, ?$

The answer is 6084.

2 solutions

Yuri Lombardo
Dec 10, 2015

No need to do all that stuff.

If we expand the series, we can see that $12^{2}$ appears exactly 12 times (as a sum), $11^{2}$ appears 11 times and it keeps going in this way.

So we can write: $1^{2}\cdot1+ 2^{2}\cdot2+...+ 11^{2} \cdot11+ 12^{2}\cdot12$ Or in the seguent way (easier and plain): $1^{3}+2^{3}+...+11^{3}+12^{3}$ That is the sum of the first 12 cubes

Now use the generalized formula to calculate the sum of the first $n$ cubes and put $n=12$ $\ \sum_{k=1}^n k^{3} = \frac{n^{2}\cdot(n+1)^{2}}{4}=\frac{12^{2}\cdot(13)^{2}}{4}= \boxed{6084}$

GJ, same way

Adrianus Felix - 5 years, 6 months ago

Did the same way. Cant comment on your formatting since I never write solutions

Shreyash Rai - 5 years, 6 months ago

Small correction in the solution. In the explanation, it should be $12^2$ and $11^2$ , not $12^{12}$ and $11^{11}$ .

Janardhanan Sivaramakrishnan - 5 years, 6 months ago

Pulkit Gupta
Dec 11, 2015

The series can be written as $\sum_{i=1}^{12}$ $n^{2}$ - $\sum_{i=1}^{12}$ $n^{2}$ ( 12 - n) = $\sum_{i=1}^{12}$ $n^{3}$ = 6084

Sum of Summations

The answer is 6084.

2 solutions

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