All Questions
Tagged with algebra-precalculus summation
977
questions
67
votes
16
answers
54k
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Proving $1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$ using induction [duplicate]
How can I prove that
$$1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$$
for all $n \in \mathbb{N}$? I am looking for a proof using mathematical induction.
Thanks
188
votes
28
answers
20k
views
Proving the identity $\sum_{k=1}^n {k^3} = \big(\sum_{k=1}^n k\big)^2$ without induction
I recently proved that
$$\sum_{k=1}^n k^3 = \left(\sum_{k=1}^n k \right)^2$$
using mathematical induction. I'm interested if there's an intuitive explanation, or even a combinatorial interpretation ...
- 5,897
118
votes
5
answers
127k
views
What is the term for a factorial type operation, but with summation instead of products?
(Pardon if this seems a bit beginner, this is my first post in math - trying to improve my knowledge while tackling Project Euler problems)
I'm aware of Sigma notation, but is there a function/name ...
- 1,419
4
votes
2
answers
2k
views
How to prove $\sum_{i=1}^n \frac1{4i^2-1}=\frac{1}{2}-\frac{1}{2 \times (2n+1)}$?
How could we show $$\sum_{i=1}^n \frac{1}{4i^2-1}=\frac{n}{2n+1}=\frac{1}{2}-\frac{1}{2 \times (2n+1)}$$ I am looking for an answer other than induction as in the actual problem I am suppose to find ...
- 22.5k
5
votes
2
answers
601
views
A "fast" way for computing $\sum\limits_{n=1}^{100} n\times 2^n $ [duplicate]
How to compute 'z', where $\displaystyle z = \sum_{n=1}^{100} n\times 2^n$ ?
The answer is of the form $99 \times 2^{101} + 2$, I need a fast approach as this problem is supposed to be solved under a ...
- 22.5k
2
votes
1
answer
407
views
Changing a single sum to a triple sum
I have the a summation of the following form:
$$\sum_{M_1} \left[
{
f(M_1-m_1,-M_1+m_1+\mu_1^\prime,\mu_1^\prime)
\cdot \atop {
\displaystyle g(M_1,-M_1+m_1+\mu_1^\prime,m_1+\mu_1^\prime)
\cdot\atop
\...
- 2,509
5
votes
1
answer
227
views
Formula for the sum $\sum_{i=2}^{n} \frac1{i^2-1}$
I have to find the formula for sum
$$\sum_{i=2}^{n} \frac1{i^2-1}$$
I remember reading somewhere that $\displaystyle \frac1{i^2-1}$ can be shown as $\displaystyle\frac1{i+1}$ and $\displaystyle \frac1{...
- 53
32
votes
5
answers
2k
views
How to prove $\sum\limits_{r=0}^n \frac{(-1)^r}{r+1}\binom{n}{r} = \frac1{n+1}$?
Other than the general inductive method,how could we show that $$\sum_{r=0}^n \frac{(-1)^r}{r+1}\binom{n}{r} = \frac1{n+1}$$
Apart from induction, I tried with Wolfram Alpha to check the validity, ...
- 22.5k
135
votes
7
answers
108k
views
Values of $\sum_{n=0}^\infty x^n$ and $\sum_{n=0}^N x^n$
Why does the following hold:
\begin{equation*}
\displaystyle \sum\limits_{n=0}^{\infty} 0.7^n=\frac{1}{1-0.7} = 10/3\quad ?
\end{equation*}
Can we generalize the above to
$\displaystyle \sum_{n=...
26
votes
1
answer
860
views
Is this algebraic identity obvious? $\sum_{i=1}^n \prod_{j\neq i} {\lambda_j\over \lambda_j-\lambda_i}=1$
If $\lambda_1,\dots,\lambda_n$ are distinct positive real numbers, then
$$\sum_{i=1}^n \prod_{j\neq i} {\lambda_j\over \lambda_j-\lambda_i}=1.$$
This identity follows from a probability calculation ...
user940
24
votes
5
answers
10k
views
Algebraic Proof that $\sum\limits_{i=0}^n \binom{n}{i}=2^n$
I'm well aware of the combinatorial variant of the proof, i.e. noting that each formula is a different representation for the number of subsets of a set of $n$ elements. I'm curious if there's a ...
- 15.5k
1
vote
3
answers
4k
views
The sum of the reciprocal of the triangular numbers up to $\frac1{t_n}$ is < 2
I am supposed to prove that $(1/1) + (1/3) + (1/6) + \dots + (1/t_n) < 2$.
The hint is that
$$
\frac2{n(n+1)} = 2\left(\frac1{n} - \frac1{n+1}\right)
$$
However, I was thinking that if you ...
- 453
6
votes
7
answers
27k
views
Proof of the formula $1+x+x^2+x^3+ \cdots +x^n =\frac{x^{n+1}-1}{x-1}$ [duplicate]
Possible Duplicate:
Value of $\sum x^n$
Proof to the formula
$$1+x+x^2+x^3+\cdots+x^n = \frac{x^{n+1}-1}{x-1}.$$
- 81
18
votes
12
answers
17k
views
How to prove this binomial identity $\sum_{r=0}^n {r {n \choose r}} = n2^{n-1}$?
I am trying to prove this binomial identity $\displaystyle\sum_{r=0}^n {r {n \choose r}} = n2^{n-1}$ but am not able to think something except induction,which is of-course not necessary (I think) here,...
- 22.5k
30
votes
5
answers
13k
views
Is there any formula for the series $1 + \frac12 + \frac13 + \cdots + \frac 1 n = ?$ [closed]
Is there any formula for this series?
$$1 + \frac12 + \frac13 + \cdots + \frac 1 n .$$
- 459