How Do I Find An Expression For The Summation Of 1^k 2^k . N^k

What's an explicit expression for either of the two sums in [math]\sum_k{\binom{m}{k} \binom{n+k}{m}} = \sum_k{\binom{m}{k} \binom{n}{k} 2^k}[/math]?

Mathematica will tell you that this sum is a case of the Hypergeometric function [math]{}_2F_1[/math] (see http://en.wikipedia.org/wiki/Hyp... ):In[1]:= Sum[Binomial[m,k] Binomial[n,k] 2^k, {k,0,m}]

Out[1]= Hypergeometric2F1[-m, -n, 1, 2]
If you’re looking for a form in terms of “simpler” functions, you’re probably out of luck.

What's the summation of k.(a^k) for k = 1 to n?

First of all, (a^k) for k = 1 to n, forms a GP with common ratio ‘a’.First term , aSecond Term, (a^2)Third Term, (a^3)… & So on till (a^n).So we can say, Σ(a^k) means sum of n terms of a GP with common ratio ‘a’.We know, Sum of n terms of a GP is given by -Σ(a^k) = a [(1-a^n)]/[1-a]; where k = 1 to n. ———-(1)People often do the mistake of multiplying ‘k’ in the above eqn(1)Which yieldskΣ(a^k) = k*a [(1-a^n)]/[1-a];This is not correct as we need the value of Σk(a^k) and not kΣ(a^k). {Mind the ‘k’ multiplication is inside the summation sign and not outside it}In order to obtain the desired value, differentiate Eqn(1) with respect to ‘a’.P.S : ‘k’ is variable with respect to incremental value, which is integer/discrete. But ‘a’ is the variable with respect to differentiation operation.On differentiation, we get -Σk*a^(k-1) = [ a/(1-a) * (-n)* a^(n-1) ] + [ (1-a^n) /(1-a)^2 ]Σk*a^(k-1) = (1-a^n)/(1-a)^2 - n(a^n)/(1-a)Σk*a^(k-1) = [(1-(a^n)) - n(1-a)(a^n)] / (1-a)^2Now multiplying both sides by ‘a’. Since incremental variable is ‘k’,multiplying both sides by ‘a’, will not alter the equation. Hence, ‘a’can be sent inside the Σ.a*Σk*a^(k-1) = Σa*k*a^(k-1) = Σk*(a^k) =a* [(1-(a^n)) - n(1-a)(a^n)] / (1-a)^2Therefore,Σk*(a^k) = a* [(1-(a^n)) - n(1-a)(a^n)] / (1-a)^2Σk*(a^k) = [a*(1-(a^n)) - n(1-a)*a^(n+1)] / [(1-a)^2]Hope this helped. Cheers!!

How can I find the closed form of the sum: [math]\sum\limits_{k=1}^{n} \binom{n}{k}k^{2} [/math]?

Let’s find a combinatorial proof. Suppose we select a committee of [math]k[/math] people out of [math]n[/math] people, which can be done in [math]{{n}\choose{k}}[/math] ways. We now choose a president and a treasurer for the committee, which are allowed to be the same person. There are [math]k^2[/math] ways to pick them both, for a total of [math]{{n}\choose{k}}k^2[/math] ways. Letting [math]k[/math] range from [math]0[/math] to [math]n[/math] gives us the sum we’re working with.To count this in another way, we consider two cases: The president and the treasurer may be the same person or different people.They’re the same. We have [math]n[/math] choices for who the president/treasurer is; she must be on the committee. There are [math]n-1[/math] other people; each of them may be in the committee or not in the committee, so we have [math]2[/math] choices to make [math]n-1[/math] times. This gives us a total of [math]n2^{n-1}[/math] for this case.They’re different. There are [math]n(n-1)[/math] choices for who the president and treasurer are; choose the president, then choose the treasurer from the remaining people. Both these people are on the committee by default. As before, there are [math]2^{n-2}[/math] ways to place the other people on the committee, for a total of [math]n(n-1)2^{n-2}[/math] ways here.Summing our cases gives a grand total of [math]n(n+1)2^{n-2} = {{n+1}\choose{2}}2^{n-1}[/math].

How do I find the closed form of the sum [math]\sum_{k=1}^{n} (2k+1) \binom{n}{k}[/math]?

By the Binomial theorem :[math] (1+x^2)^n=\sum_{k=0}^{n}\binom{n}{k}x^{2k} [/math]Hence, [math]x(1+x^2)^n=\sum_{k=0}^{n}\binom{n}{k}x^{2k+1} [/math]Differentiating both sides w.r.t. [math] x[/math], we have the following identity [math] 2nx(1+x^2)^{n-1} + (1+x^2)^n [/math][math]= \sum_{k=0}^{n} (2k+1)\binom{n}{k} x^{2k} [/math]. Finally, putting [math]x=1[/math] on both sides of this identity and transposing the term corresponding to [math]k=0[/math], we conclude that [math]\sum_{k=1}^{n} (2k+1)\binom{n}{k} = (1+n)2^n-1[/math]

How do I show that [math]\sum\limits_{n=k}^{\infty}\frac{1}{n^2}<\frac{2}{k}[/math]?

Sometimes, we can approximate the area under a curve by approximating it with rectangles. At other times, we can do the opposite, and approximate a sum by computing the area under a curve.The figure below shows how we might approximate the sum[math] \sum_{n=2}^{10} 1/n^2 [/math]. The upper bound is [math] \int_1^{10} 1/x^2 dx [/math]. (Created from a visualization on the Mathematica website: Visualize Riemann Sums.)Note that, since we are using right endpoints for our rectangles, the lower limit of our integral is one less than the lower limit of our sum. This will cause a slight problem when the lower limit of our sum is 1, which I will deal with later. Assume that's not the case for now.To solve this problem, we'll compute the area under the curve [math] y = 1/x^2, [/math] for [math] k - 1 \leq x \leq \infty. [/math] Technically, this is an improper integral, and should be computed with a limit. But I'm going to be lazy.[math] \int_{k - 1}^{\infty} 1/x^2 dx [/math][math] = [-1/x]_{k - 1}^{\infty} = 1/(k - 1). [/math]Now if [math] k \geq 2 [/math], then [math] 1/(k - 1) \leq 2/k [/math]. So we have shown that our inequality holds for every case with [math] k \geq 2 [/math].We now just have to deal with the troublesome case of when k = 1. To get a better approximation with the integral method, we can always sum up some number of terms in our series, and then use an integral to approximate the rest.In this case we only have to add the first term.[math] \sum_{n = 1}^{\infty} 1/n^2  [/math][math] = 1 + \sum_{n = 2}^{\infty} 1/n^2 [/math][math] \leq 1 + \int_{x = 1}^{\infty} 1/x^2 dx = 2 [/math]So when k = 1, our sum is indeed bounded by 2/k = 2.QED.