Linked Questions

2
votes
2answers
2k views

Partial Sum of a Polynomial [duplicate]

Possible Duplicate: why is $\sum\limits_{k=1}^{n} k^m$ a polynomial with degree $m+1$ in $n$ I'm searching for a way to find the partial sum of a polynomial, is there any way of doing this ...
1
vote
4answers
448 views

Power summation of $n^3$ or higher [duplicate]

Possible Duplicate: why is $\sum\limits_{k=1}^{n} k^m$ a polynomial with degree $m+1$ in $n$ If I want to find the formula for $$\sum_{k=1}^n k^2$$ I would do the next: $$(n+1)^2 = a(n+1)^3 + b(n+...
2
votes
3answers
460 views

Evaluating N-th partial sums of polynomials [duplicate]

Possible Duplicate: why is $\sum\limits_{k=1}^{n} k^m$ a polynomial with degree $m+1$ in $n$ How can I find $\sum_{n=1}^N n^2-n$? Wolfram Alpha will tell you that it is $\frac{N}{3} (N-1)(N+1)$, ...
1
vote
3answers
161 views

Finite summation [duplicate]

Possible Duplicate: why is $\sum\limits_{k=1}^{n} k^m$ a polynomial with degree $m+1$ in $n$ What is the proof without induction for : $(1)$ $\sum_{i=1}^n\ i^2= \frac{n(n+1)(2n+1)}{6}$ $(2)$ $\...
1
vote
1answer
210 views

Formula for $\sum_{k=1}^{n}{k^p}$ where p is a positive integer [duplicate]

Possible Duplicate: why is $\sum\limits_{k=1}^{n} k^m$ a polynomial with degree $m+1$ in $n$ Any hints that can take me from here or am I completely lost. $\sum_{k=1}^{n}{k^p}=\sum_{a=1}^{p}(-1)^...
0
votes
2answers
94 views

Finding an expression for the sum of n tems of the series $1^2 + 2^2 + 3^2 + … + n^2$ [duplicate]

Possible Duplicate: why is $\sum\limits_{k=1}^{n} k^m$ a polynomial with degree $m+1$ in $n$ I know that if you have a non-arithmetic or geometric progression, you can find a sum $S$ of a series ...
0
votes
0answers
79 views

Sum of series: $1^k+2^k+3^k+…+n^k =?$ [duplicate]

Is there any better algorithm to solve this equation other than brute force. $1^k+2^k+3^k+...+n^k=$ formula? Here $k$ is a natural number, $n$ is a natural number.
118
votes
31answers
61k views

Sum of First $n$ Squares Equals $\frac{n(n+1)(2n+1)}{6}$

I am just starting into calculus and I have a question about the following statement I encountered while learning about definite integrals: $$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$ I really ...
7
votes
4answers
376 views

Formula for $1^2+2^2+3^2+…+n^2$

In example to get formula for $1^2+2^2+3^2+...+n^2$ they express $f(n)$ as: $$f(n)=an^3+bn^2+cn+d$$ also known that $f(0)=0$, $f(1)=1$, $f(2)=5$ and $f(3)=14$ Then this values are inserted into ...
2
votes
1answer
6k views

How to calculate $ 1^k+2^k+3^k+\cdots+N^k $ with given values of $N$ and $k$? [duplicate]

Here $ 1<N<10^9$ and $0<k<50$ So we have to calculate it in order of $O(\log N)$.
2
votes
2answers
967 views

Conceptual basis for formula for sum of n first integers raised to power k,

Most programmers (including me) are painfully aware of quadratic behavior resulting from a loop that internally performs 1, 2, 3, 4, 5 and so on operations per iteration, $$\sum_{i=1}^n i = \frac{n \...
1
vote
3answers
601 views

Find the value of a succession of additions

$$1^2+2^2+3^2+...+10000$$ How do you find the exact value of that? I'm studying induction, and I'm still not sure how to get that value.
1
vote
6answers
128 views

Find $\lim_{n\to\infty} \frac{1^4+2^4+\dots+n^4}{1^4+2^4+\dots+n^4+(n+1)^4}$ [closed]

I am just trying to calculate $$\lim_{n\to\infty} \frac{1^4+2^4+\dots+n^4}{1^4+2^4+\dots+n^4+(n+1)^4}.$$ To do this I apply formula for sum of fourth powers of $n$ number. My result: $$\lim_{n\to\...
1
vote
2answers
203 views

Identity with Bernoulli numbers: $\sum\limits_{k=1}^{n}k^p=\frac{1}{p+1}\sum\limits_{j=0}^{p}\binom{p+1}{j}B_j n^{p+1-j}$

How I can prove that $$\sum_{k=1}^{n}k^p=\frac{1}{p+1}\sum_{j=0}^{p}\binom{p+1}{j}B_j n^{p+1-j},$$ where $B_j$ is the $j$th Bernoulli number? I hope to find the answer. Thanks for help.
2
votes
6answers
94 views

Find the sum to n terms

$$S=1^2+3^2+6^2+10^2+15^2+.......$$ My attempt is as follows: $$T_n=\left(\frac{n\cdot\left(n+1\right))}{2}\right)^2$$ $$T_n=\frac{n^4+n^2+2\cdot n^3}{4}$$ $$S=\frac{1}{4}\cdot\sum_{n=1}^{n}\left(...

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