All Questions
Tagged with algebra-precalculus summation
977
questions
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
141
votes
36
answers
308k
views
Proof that $1+2+3+4+\cdots+n = \frac{n\times(n+1)}2$
Why is $1+2+3+4+\ldots+n = \dfrac{n\times(n+1)}2$ $\space$ ?
- 1,585
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=...
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
78
votes
8
answers
12k
views
Multiple-choice: sum of primes below $1000$
I sat an exam 2 months ago and the question paper contains the problem:
Given that there are $168$ primes below $1000$. Then the sum of all primes
below 1000 is
(a) $11555$ (b) $76127$ (c) $...
- 3,789
67
votes
16
answers
54k
views
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
56
votes
7
answers
8k
views
Can we calculate $ i\sqrt { i\sqrt { i\sqrt { \cdots } } }$?
It might be obvious that $2\sqrt { 2\sqrt { 2\sqrt { 2\sqrt { 2\sqrt { 2\sqrt { \cdots } } } } } } $ equals $4.$ So what about $i\sqrt { i\sqrt { i\sqrt { i\sqrt { i\sqrt { i\sqrt { \cdots } } } } } } ...
user536517
41
votes
5
answers
21k
views
How to find the sum of the series $1+\frac{1}{2}+ \frac{1}{3}+\frac{1}{4}+\dots+\frac{1}{n}$?
How to find the sum of the following series?
$$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{n}$$
This is a harmonic progression. So, is the following formula correct?
$\frac{(number ~...
- 10k
39
votes
5
answers
76k
views
Simple Double Summation
I've seen how nesting works with a simple $(i+j)$ but this problem below is tripping me up. It's either because of the multipliers or because they each start at zero but I get 60, and the answer I ...
- 661
35
votes
3
answers
4k
views
Trig sum: $\tan ^21^\circ+\tan ^22^\circ+\cdots+\tan^2 89^\circ = \text{?}$
As the title suggests, I'm trying to find the sum $$\tan^21^\circ+\tan^2 2^\circ+\cdots+\tan^2 89^\circ$$
I'm looking for a solution that doesn't involve complex numbers, or any other advanced branch ...
- 3,153
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
31
votes
2
answers
77k
views
What is the square of summation?
Consider the following, which one of the following is true ??
$$\left( \sum^{n-1}_{j=0}Z_j\right)^2 = \sum^{n-1}_{j=0} Z_j^2 + \sum^{n-1}_{j\neq i} Z_i Z_j$$
OR
$$\left( \sum^{n-1}_{j=0}Z_j\right)^...
- 1,237
30
votes
5
answers
13k
views
Is there any formula for the series $1 + \frac12 + \frac13 + \cdots + \frac 1 n = ?$ [closed]
Is there any formula for this series?
$$1 + \frac12 + \frac13 + \cdots + \frac 1 n .$$
- 459
29
votes
2
answers
829
views
How to prove $\sum_{n=0}^{\infty} \frac{1}{1+n^2} = \frac{\pi+1}{2}+\frac{\pi}{e^{2\pi}-1}$
How can we prove the following
$$\sum_{n=0}^{\infty} \dfrac{1}{1+n^2} = \dfrac{\pi+1}{2}+\dfrac{\pi}{e^{2\pi}-1}$$
I tried using partial fraction and the famous result $$\sum_{n=0}^{\infty} \...
user208998
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