$$S= 1\times(1) + 2\times(1+2) + 3\times(1+2+3) +\dots+ n\times(1+2+3+\dots+n)$$
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1$S =\sum \frac{n(n+1)}{2}n$. – Dec 28 '16 at 15:44
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Hint: $1+2+\dots+n=\dfrac{n(n+1)}{2}$. – Galc127 Dec 28 '16 at 15:46
2 Answers
5
We need three formulas. The sum of the $n$ first numbers, the sum of the $n$ first squares and the sum of the $n$ first cubes:
$$\sum_{j=1}^n j=\frac{n(n+1)}2$$ $$\sum_{j=1}^n j^2=\frac{n(n+1)(2n+1)}6$$ $$\sum_{j=1}^n j^3=\frac{n^2(n+1)^2}4$$
Now, $$\sum_{j=1}^nj\sum_{k=1}^jk=\sum_{j=1}^n\frac{j^2(j+1)}2=\frac12\left(\sum_{j=1}^nj^3+\sum_{j=1}^nj^2\right)=\frac{n(n+1)(2n+1)}{12}+\frac{n^2(n+1)^2}8$$
hence $$S=\frac{n(n+1)(n+2)(3n+1)}{24}$$
2
First, notice that $1+2+3+\dots+n=\frac{n(n+1)}2$
Then notice that
$$\sum_{k=1}^nk\frac{k(k+1)}2=\frac12\sum_{k=1}^nk^3+k^2=\frac12\left(\left(\frac{n(n+1)}2\right)^2+\frac{n(n+1)(2n+1)}6\right)$$
hence $$S=\frac{n(n+1)(n+2)(3n+1)}{24}$$
For more information as to how to derive these formulas, see here.
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