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.

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calculation of summing the product of combinations

$$\sum_{i=0}^{10}\sum_{i=j}^{10}{}^{10}C_j {}^{j}C_i $$ I tried to expand the terms ,but I can't put sum them to get a legit answer. ${}^{10}C_0 {}^{0}C_0+{}^{10}C_1 {}^{1}C_0.... +{}^{10}C_1 {}^{1}C_1 +{}^{10}C_2 {}^{2}C_1+..........…
Nebo Alex
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Simplifying the generalization of a summation

I guess I am having trouble with the algebra, could someone walk me through this probably simple simplification? I currently have $\frac{x^2\sqrt3}{4} + $$\sum_{i=0}^n 3\cdot4^i \frac{\frac{x}{3^{i-1}}^2 \sqrt3}{4} $ I am pretty sure that I am…
Cade
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Is it possible to make $\sum_{k=0}^{n-i}10^k \binom{n-i}{k}$ shorter?

I've got some expression $$\sum_{i=1}^n\left( 2^{i-1} a_i \sum_{k=0}^{n-i}10^k \binom{n-i}{k} \right)$$ $a_i$ stands for i-th digits in number. I came up with this formula from task in which we had to delete some set of digits from given number…
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How to write this equation in terms of Sigma

I am sure this is trivial to most, but wanted to confirm that this is how you write the sum for this sequence: $$a_{11}b_{11}+a_{12}b_{21}+a_{13}b_{31}+a_{14}b_{41}+\cdots+a_{1j}b_{i1}=\sum_{n=1}^j\sum_{m=1}^ia_{1n}b_{m1}$$
HugHes
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Evaluating $\displaystyle\sum_{k=1}^{2017}\dfrac 1{n^2+2n}$

I tried simplifying it by first resolving it into partial fractions but to no use. Also I know that $\sum (1/n)$ is harmonic number, but I don't know how to use it. Please help me.
Yami Kanashi
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Double Summation Simplification - How to simplify the below summation formula with powers of 2?

I need help in simplifying the below double summation. $\sum_{j=1}^i \sum_{k=i}^n (k-j+1) (2^{\max(j-2,0)})(2^{\max(n-k-1,0)})$
Nilanshu96
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Summation limits change

I know this is a very naive question, but I still struggle with summation limits. How does someone go from the first summation to the second? I am looking at the second equation, second step. We want to make the index start at $0$ instead of $-m$,…
Euler_Salter
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How would I derive the partial sum formula for $\sum _{ n=1 }^{ \infty }{ (-1)^{ n }\cdot n } $?

I was solving a practice competitive programming question when I realized that this seems very familiar to the things I have done in a Calculus class. So, I tried to figure out the closed form partial sum for $\sum _{ n=1 }^{ m }{ (-1)^{ n }\cdot n…
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why this is true when proofing euler summation rule?

In this proof of euler's summation,After derivation it said that : Second thing he assumed was: $$\sum_{k=0}^{\infty} (k+1)x^k=1-2+3-4+5-6+...=\frac{1}{4}$$ & I want to know how we can get to this . Thanks
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How would I get the value of this sum in terms of n

How would i evaluate this sum as an nth term: $\sum_{r=1}^{n} r!r$ Also is there any particular method for simplifying sums that have an nth term with a factorial in?
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How to evaluate $\sum_{n=1}^{\infty}\frac{1}{n}(\frac{1}{2^n}-\frac{1}{4^n})$

I want to calculate the infinite summation below : $$\sum_{n=1}^{\infty}\frac{1}{n}(\frac{1}{2^n}-\frac{1}{4^n})$$ but I totally failed to approach. Thanks for any help. Wolframalpha gave me the results : $\ln(3/2)$.
Jinmu You
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Very remedial calculus.

It has been awhile, and I know this is very basic, but if someone would help me just get this, I know I can start retraining my brain. Basically it is this. I have : number of digits x position of digit y value of digit z so, iterating through the…
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How to reduce this double summation in to a single summation?

So I was participating in this contest and now that it's over I was checking the editorial for the author solution. Basically the solution was deriving a single summation from a double summation in order to have a more efficient solution .. But I…
motatoes
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Compute the sum of binomial probability

I have a sum that is $$\sum_{k=\lfloor{\frac{n}{2}+1}\rfloor}^n\binom{n}{k}p^n$$ I try $n=3$ and $n=4$, I got $$\sum_{k=\lfloor{\frac{3}{2}+1}\rfloor}^3\binom{n}{k}p^3=4p^3$$ $$\sum_{k=\lfloor{\frac{4}{2}+1}\rfloor}^4\binom{n}{k}p^4=5p^4$$ How do I…
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Open interval by summation

$A = \{\sum_{k=1}^{\infty} \frac{a_k}{5^k}\}$ where $ a_k = [0,1,2,3,4]$ can the set $A$ contain an open interval? How?
jnyan
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