Mathematics Stack Exchange
Mathematics Stack Exchange

Questions tagged [summation]

Questions about evaluating summations, especially finite summations. For infinite series, please consider the (sequences-and-series) tag instead.

The notation $\sum\limits_{i=1}^na_i$ means $a_1+\ldots +a_n$.

Use for sums of infinite series and questions of convergence; use for questions about finite sums and simplification of expressions involving sums.

17770 questions
3
votes
1 answer

Help getting a sum into another form

I have the following sum: $$\sum_{i=0}^n\sum_{j=0}^{n-i}a_{i,j}r^is^jt^{n-i-j}u^{2n-2i-j},$$ which I'd like to get into the form $$\sum_{k=0}^{2n}\lambda_ku^k.$$ In the case $n=3$ I have formed the following analysis of the indices…
pshmath0
  • 10,565
3
votes
2 answers

Summation and fractions

I'm trying to solve the following summation: $$ n \sum_{i=0}^{\log_3 (n-1)} \log_3 \frac{n}{3^i} $$ The last part is $\log (\frac{n}{3^i})$. I can't by the life of me figure this out. I know that it should equal this: $$ \sum_{i=0}^{\log_3n-1}…
Garry
  • 33
  • 3
3
votes
3 answers

Is there a quick way to see these sums are equal?

I can show that these two sums are equal by writing out all terms, but is there a quick way to see the equality: $$\sum_{i=0}^n\sum_{j=0}^i a_jb_{i-j}c_{n-i} = \sum_{i=0}^n\sum_{j=0}^{n-i} a_ib_{j}c_{n-i-j} $$ (this question popped up while checking…
user370967
3
votes
2 answers

Find the sum $\sum\limits_{i=1}^{n}i^k$ using integration.

Observe that we can find the sum $\sum\limits_{i=1}^{n}i^3$ using "integration": $$\sum\limits_{i=1}^{n}\frac{i^3}{3}=\int\sum\limits_{i=1}^{n}i^2dn.$$ and actually we get exactly the right result(if we know the sum of $i^2$ in terms of $n$) For…
William Sun
  • 2,463
  • 9
  • 17
3
votes
6 answers

What is the value of $ \sum _{k=1}^n\frac{x_k}{x_k-1}$?

What is the value of $$\sum _{k=1}^n\frac{x_k}{x_k-1}$$ given that $x_1,x_2,\dots,x_n $ are the roots of the equation $x^n-3x^{n-1}+2x+1=0\,$? I wrote it as $ n+\sum _{k=1}^n\frac{1}{x_k-1}$ but didn't really help much. I think Vieta's…
Alexander
  • 563
3
votes
1 answer

Power Sum of Integers and Relationship with Sum of Squares and Sum of Cubes

Let $\displaystyle\sigma_m=\sum_{r=1}^n r^m$. Refer to the tabulation of the power sum of integers here. It is interesting to note that $$\begin{align} \color{green}{\sigma_1}\ &=\frac 12 n(n+1)\\ \color{blue}{\sigma_2}\ &=\frac 16…
Hypergeometricx
  • 22,657
3
votes
3 answers

Finding sum $\sum_{i=1}^n \frac1{4i^2-1}$

I have been having problem with calculating the following summation: $$ \sum_{i=1}^n {1\over 4i^2-1} = {1\over3} + {1\over15} + {1\over35} + \cdots + {1\over 4n^2-1} $$ I do know the answer, but just can not find the way to get it. Thank you in…
user48512
  • 87
3
votes
0 answers

Finite rational sum

By a direct calculation, it is easy to deduce that the simple finite sum $$\sum\limits_{j = 1}^n {\frac{{{r^2}}} {{{j^2}r + j{r^2}}}} = {H_n} + {H_r} - {H_{n + r}},$$ where $H_n$ denotes the harmonic number defined by $${H_n} = \sum\limits_{j=…
xuce1234
  • 1,330
  • 7
  • 12
3
votes
1 answer

Find the sum of the following series: $ \frac{1}{1 \cdot 2}\,+\,\frac{1}{2 \cdot 3}\,+\,\frac{1}{3 \cdot 4} +\cdots\,+\frac{1}{100 \cdot 101}$

Find the sum of the following series $$ \frac{1}{1 \cdot 2}\,+\,\frac{1}{2 \cdot 3}\,+\,\frac{1}{3 \cdot 4} +\cdots\,+\frac{1}{100 \cdot 101}$$ My Attempt: $$\frac{1}{2 \cdot 3} =\frac{1}{2}-\frac{1}{3}.$$ So we can write question…
laura
  • 2,530
3
votes
2 answers

$1^2$+$2^2$-$2×3^2$+$4^2$+$5^2$-$2×6^2$+$...$+$9997^2$+$9998^2$-$2×9999^2$

∵ For any positive integer $n$ $n^2-2(n+1)^2+(n+2)^2$ = $n^2-2n^2-4n-2+n^2+4n+4=2$ ∵ $1^2$+$2^2$-$2×3^2$+$4^2$+$5^2$-$2×6^2$+$...$+$9997^2$+$9998^2$-$2×9999^2$ = $1^2+(2^2-2×3^2+4^2)+(5^2-2×6^2+7^2)+...+(9995^2-2×9996^2+9997^2)+9998^2-2×9999^2$ =…
Chen Mo
  • 137
3
votes
1 answer

Number of powers of $2$ having leading digit $1$

How many of the numbers $2^m$ (where $0\le m\le M)$ have leading digit $1$? My trial - Since leading digit $=1$, whenever $2^m$ reaches or just crosses a $10^x$ and is less than $2 \cdot 10 ^ x$, that $m$ is eligible i.e. $$10^x \le 2^m \lt 2\cdot…
Andariel
  • 159
3
votes
1 answer

Checking computation of $\sum\limits_{i=1}^{n-1}\sum\limits_{j=i+1}^{n} \frac12$

$$\sum_{i=1}^{n-1}\sum_{j=i+1}^{n} 1/2 = \sum_{i=1}^{n-1} (n-1) (1/2) $$ We remove the first summation because it just sums all $1/2$ in $n-1$ times. So we get $(n-1)(1/2)$. continue: $$ \sum_{i=1}^{n-1} (n-1) (1/2) = (n-1)(n-1)(1/2)$$ Is this…
YOUSEFY
  • 639
  • 9
  • 21
3
votes
2 answers

Proving $ 2n^{n-3} = \sum\limits_{i=1}^{n-1}\binom{n-2}{i-1}i^{\ i-2}(n-i)^{\ n-i-2} $

My question is how to show$$ 2n^{n-3} = \sum_{i=1}^{n-1}\binom{n-2}{i-1}i^{\ i-2}(n-i)^{\ n-i-2} $$ I got to this result through a problem of counting labeled trees, but when I try to approach actually computing the sum, I've had no breakthroughs.
mshlis
  • 213
  • 1
  • 5
3
votes
2 answers

Prove that $ \sum_{x=1}^ \infty \frac{1}{x!} = e -1 $

I was playing around with factorials and integrals when I put $ \sum_{x=1}^ \infty \frac{1}{x!} $ in Wolfram Alpha I got the result as equal to $e - 1$. It was an amazing discovery for me. I tried to prove it, but could go nowhere. I just know…
user399078
3
votes
2 answers

Calculating with summations

$$b! \left (\sum_{n=0}^{\infty}\frac{1}{n!}-\sum_{n=0}^{b}\frac{1}{n!} \right )=\sum_{n=b+1}^{\infty}\frac{b!}{n!}> 0$$ I don't understand why $n=b+1$ in the last step of the expression.
Andreas
  • 1,948
  • 2
  • 16
  • 45
1 2 3
…
99 100