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
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Closed form summation with $n-1$ on top.

Why does the close form of the summation $$\sum_{i=0}^{n-1} 1$$ equals $n$ instead of $n-1$?
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Evaluate the Sum

Evaluate $$\sum_{n=0}^{\infty}\frac{5n+1}{(2n+1)!}$$ I tried using $e^{x}=\sum_{k=0}^{\infty}\frac{x^k}{k!}$ But I got $$5e-8\sum_{n=0}^{\infty}\frac{1}{2n+1}$$ Hints would be appreciated!
Ajay
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How does $\sum x_i y_i = \bar{x}\sum y_i$?

This is the identity given in my textbook with regards to the centre of mass of of a set of particles along the line. $x$ could be the distance along the line, and $y$ could be the mass. $\sum x_i y_i$ would then be the sum of moments of forces.
Mihail
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Summation involving the index multiplied by an exponential?

$$\begin{align} \sum_{i=0}^{n-1}ia^i &= 0 + a + 2a^2 + \cdots + (n-1)a^{n-1}\\ &= \frac{a - na^n + (n-1)a^{n+1}}{(1 - a)^2} \end{align}$$ Is there any formula for a slightly similar one? $$ \sum_{i=0}^{n-1}ia^{n-1-i} = 0 + a^{n-2} + 2a^{n-3} +…
user34295
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Please help with infinite sum with more than one variable

I built this sum for an algorithm, but am having trouble figuring out where to start solving it. If I could get it down to a partial sum, I could use limits to solve it, but its been so long since I've done anything other than the most basic…
Chris
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Alternate Sum of Cubes Using Identity

By using $x^3-(x-y)^3 = y(3x^2-3xy+y^2)$, find the value of $$(1^3-2^3+3^3-\cdots+9^3)\times3 + (10^3-11^3+12^3-\cdots-19^3).$$ I find different solutions regarding the alternate sum of cubes here, but none of the solutions shows a clear usage…
Nighty
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How to calculate a special sum?

Thank you for giving me an idea to try to find a way to calculate this sum: $$S= \sum_{i=0}^n \lfloor\sqrt{i} \rfloor$$
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How to write an expression using sigma notations?

If I have $4^{n-1}*6+4^{n-2}*6+4^{n-3}*6+...+4*6 + 6$ how do I rewrite such expression as sum notation? I do know the answer, I don't really understand how to go from this expression and make a sigma notation?
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Geometric Series with fractions (summation)

I am having trouble trying to solve this summation (with fraction) $$f(n) = \sum\limits_{i=1}^{n} \frac{-5}{6^i}$$ How do I solve this?
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How would I write this as a Summation?

I have a function n that I want to write a summation for. How would I write this as a sum? $ (1/n_1)+((1/n_2)*(1/n_1))+((1/n_3)*(1/n_2)*(1/n_1))+ ... $ etc.
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Expectation value summation

I’m running into problems with a summation $$\langle p \rangle = \frac{\sum_{p=0}^{N-1}px^p}{\sum_{p=0}^{N-1}x^p} = x\frac{d}{dx}ln(\sum_{p=0}^{N-1}x^p)$$ I know I’ve seen this before but I can’t remember the justification to get it summed…
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How to approach $\sum_{i=1}^m e^{\zeta_i}$ ($\zeta_i\leq0$)

I would like to estimate (from above) the following sum $$\sum_{i=1}^m e^{\zeta_i}\quad \quad \zeta_i\leq0$$ The problem is that I do not know $\zeta_i$! (But I know $\sum_{i=1}^m \zeta_i$...) How would you handle this problem?
Mark
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How does $e^{-q} \sum_k \frac{q^k}{(k-1)!}$ become $qe^{-q} \sum_k \frac{q^k}{k!}$?

$$e^{-q} \sum_k \frac{q^k}{(k-1)!}$$ The next step is $$qe^{-q} \sum_k \frac{q^k}{k!}$$ Where does the $q$ come from?
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$\sum_{k=0}^{\infty} \sum_{n=0}^{\infty} \frac{(n+k)!}{n! k!} x^n = \sum_{k=0}^{\infty} (1-x)^{-k-1}$

I am having trouble with proving following equalities \begin{align} \sum_{k=0}^{\infty} \sum_{n=0}^{\infty} \frac{(n+k)!}{n! k!} x^n = \sum_{k=0}^{\infty} (1-x)^{-k-1} \end{align} How this equalities holds?
phy_math
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Simplifying $(3/2)^i$ for $i = 0$ to $\\(log_2 n) -1$

$$\sum_{i=0}^{\\(log_2 n)-1} \frac{3^i}{2^i}$$ I have been working on solving a recurrence, and finally simplified to this step. However, I am lost as to where to proceed now. I know the sum of powers of 2, but I don't know what to do since it's in…