Finit Sum $\sum\limits_{i=1}^{100}i^8-2 i^2$

Question

Can anyone help me? How can I find $$\sum_{i=1}^{100}i^8-2i^2 $$

Answer 1

Perhaps this Wikipedia entry might come in useful ?

Answer 2

First of all $$k^5-2k^2=k^{\underline5}+10k^{\underline4}+25k^{\underline3}+13k^{\underline2}-k^{\underline 1},$$ where $$k^{\underline n}=k(k-1)\cdots(k+1-n).$$ Since $$(k+1)^{\underline{n+1}}-k^{\underline{n+1}}=(k+1)k^{\underline n}-k^{\underline n}(k-n) =(n+1)k^{\underline n},$$ we have $$\sum_{k=a}^b k^{\underline n}=\frac1{n+1}\sum_{k=a}^b (k+1)^{\underline{n+1}}-k^{\underline{n+1}}=\frac{(b+1)^{\underline{n+1}}-a^{\underline{n+1}}}{n+1}.$$ Therefore (using $1^{\underline n}=0$ for $n>1$) $$\sum_{k=1}^{100}k^5-2k^2=\frac16 101^{\underline 6}+2\cdot 101^{\underline 5}+\frac{25}4 101^{\underline4}+\frac{13}3 101^{\underline 3}-\frac12 101^{\underline2} .$$

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Oh, how did I arrive at the first equation? Actually by cheating, but what you can do is note that $k^{\underline n}=0$ for $k=0,1,\ldots,n-1$ and make sure that you choose the coefficients on the right hand side such that both polynomials agree for $k=0,\dots,5$. So $0^5-0^2=0$, so RHS does not need a constant term. Now $1^5-1^2=-1$, so RHS should have $-k^{\underline1}$. Next $2^5-2\cdot2^2=24$, $-2^{\underline 1}=-2$, so we need 26 more on the RHS. But $2^{\underline 2}=2$, so we add $13k^{\underline 2}$. And so on.

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And should you think that $k^{\underline n}$ looks ugly, you can work with $\binom kn$ instead. You may already have seen the equation $$\sum_{k=1}^N\binom kn=\binom{N+1}{n+1},\qquad n>0,$$ which is easily derived from the above sum formula, but also has a combinatorial interpretation: To pick $n+1$ objects out of $N+1$, first pick the last object in position $k+1$ and then pick $n$ out of the $k$ objects before that.

Now \begin{multline*}\sum_{k=1}^{100}k^5-2k^2=\sum_{k=1}^{100}120\binom k5+240\binom k4+150\binom k3+26\binom k2-\binom k1=\\=120\binom{101}6+240\binom {101}5+150\binom{101}4+26\binom{101}3-\binom{101}2.\end{multline*}

Answer 3

Using the Euler-Maclaurin Sum Formula (which is exact for polynomials), we get $$ \begin{align} \sum_{i=1}^ni^8-2i^2 &=C+\overbrace{\left(\frac19n^9-\frac23n^3\right)}^{\int f(n)\,\mathrm{d}n}+\frac12\overbrace{\left(n^8-2n^2\right)}^{f(n)}+\frac1{12}\overbrace{\left(8n^7-4n\right)}^{f'(n)}\\ &-\frac1{720}\overbrace{\left(336n^5\right)}^{f'''(n)}+\frac1{30240}\overbrace{\left(6720n^3\right)}^{f^{(5)}(n)}-\frac1{1209600}\overbrace{\left(40320n\right)}^{f^{(7)}(n)}\\[16pt] &=\frac19n^9+\frac12n^8+\frac23n^7-\frac7{15}n^5-\frac49n^3-n^2-\frac{11}{30}n \end{align} $$ We get that $C=0$ by plugging in $n=0$.

Therefore, $$ \sum_{i=1}^{100}i^8-2i^2=116177773110656630 $$

Answer 4

Hint: Expand following formula for $k=2,5,8$ $$\sum_{i=1}^n i^k=\frac{(n+1+B)^{k+1}-B^{k+1}}{k+1}$$ $$B^0=1,B^1=\frac{-1}{2},B^2=\frac{1}{6},B^3=0,B^4=\frac{-1}{30},B^5=0,B^6=\frac{1}{42},B^7=0,B^8=\frac{-1}{30},B^9=0,....$$

Answer 5

Because: $$S_1=\sum_{k=1}^N k^5=\frac{1}{12}N^2(2N^2+2N-1)(N+1)^2$$ and: $$S_2=\sum_{k=1}^N 2k^2=\frac{1}{6}N(N+1)(2N+1)$$ you have: $$S=S_1-S_2=\frac{1}{12}N(N+1)(2N^4+4N^3+N^2-9N-4)$$ putting $N=100$ you have the result. Obviously this is valid if your question is $\sum(i^5-2i^2)$ If the question is $\sum(i^8-2i^2)$, the answer is: $$S=\frac{1}{90}N(N+1)(2N+1)(5N^6+15N^5+5N^4-15N^3-N^2+9N-33$$