Finit Sum $\sum\limits_{i=1}^{100}i^8-2 i^2$ Can anyone help me?
How can I find 
$$\sum_{i=1}^{100}i^8-2i^2 $$
 A: Perhaps this Wikipedia entry might come in useful ?
A: 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 $n>1$)
$$\sum_{k=1}^{100}k^5-2k^2=\frac16 101^{\underline 6}+2\cdot 101^{\underline 5}+\frac{25}4 101^{\underline4}+\frac{13}3 101^{\underline 3}-\frac12 101^{\underline2}
.$$

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.

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*}
A: 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,
$$
\sum_{i=1}^{100}i^8-2i^2=116177773110656630
$$
A: Hint: Expand following formula for $k=2,5,8$
$$\sum_{i=1}^n i^k=\frac{(n+1+B)^{k+1}-B^{k+1}}{k+1}$$
$$B^0=1,B^1=\frac{-1}{2},B^2=\frac{1}{6},B^3=0,B^4=\frac{-1}{30},B^5=0,B^6=\frac{1}{42},B^7=0,B^8=\frac{-1}{30},B^9=0,....$$
A: 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:
$$S=\frac{1}{90}N(N+1)(2N+1)(5N^6+15N^5+5N^4-15N^3-N^2+9N-33$$
