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$$\sum_{i=1}^n(-1)^{i+1}i(i+1)$$ I know $$\sum_{i=1}^ni(i+1)=\frac{n(n+1)(n+2)}3$$ but not for the first expression. Thank you in advance and I'm not a mathematician or math student so just ask me if any information is needed.

Dharmik Gadhiya
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2 Answers2

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I'll write the general term as $$a_n=(-1)^{n+1}n(n+1)\qquad n=1,2,\cdots$$

Then, $$\sum_{k=1}^{2h} a_k=\sum_{k=1}^{h}a_{2k}+a_{2k-1}=\sum_{k=1}^h2k(2k-1)-(2k+1)2k=-\sum_{k=1}^h4k=-2h(h+1)$$

So, for even values of $n$, $$\sum_{k=1}^n a_k=-\frac{n(n+2)}2=-\frac{n^2}2-n$$ On the other hand, if $n$ is odd, $$\sum_{k=1}^n a_k=n(n+1)=\sum_{k=1}^{n-1}a_k=n(n+1)-\frac{(n-1)(n+1)}{2}=\\=\frac{n^2}2+n+\frac12$$

One formula covering both cases $$\sum_{k=1}^na_k=\frac14+(-1)^{n+1}\frac{2n^2+4n+1}{4}$$

  • dont you think +1 should be +2 because for n = 3 it is giving 7.75 instead of 8 and for n = 5 it is giving 17.75 instead of 18. – Dharmik Gadhiya Aug 19 '17 at 09:58
  • @DharmikGadhiya $$n=3\longrightarrow\frac14+(-1)^4\frac{18+12+1}{4}=\frac{1+31}4=8\ n=5\longrightarrow\frac14 +(-1)^6\frac{50+20+1}{4}=\frac{1+71}{4}=18$$ –  Aug 19 '17 at 10:13
  • yes sure i was calculation from second last formula. – Dharmik Gadhiya Aug 20 '17 at 14:44
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Grouping in pairs

$$1\cdot2-2\cdot3+3\cdot4-4\cdot5+5\cdot6-6\cdot7+\cdots=-2\cdot2-2\cdot4-2\cdot6-\cdots\\=-4(1+2+3+4+\cdots).$$

Use the triangular numbers formula, and for an odd number of terms, add the final $n(n+1)$.