Sequences Problem

Sum the series: 121^{2} + 323^{2} + 626^{2} + 10210^{2} + 15215^{2} + ..........n terms

Please give me the shortest way to solve this problem.

Note by Nishant Sharma
6 years, 5 months ago

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10 votes

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As Aditya pointed out, it is useful to know that i=1ni4=n(n+1)(2n+1)(3n2+3n1)30. \sum_{i=1}^n i^4 = \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}. Another useful formula is that

i=1ni3=[n(n+1)2]2. \sum_{i=1}^n i^3 = \left[ \frac { n(n+1)}{2} \right ]^2.

With this, we can write the sum as

i=1n[n(n+1)2]2=i=1nj=1ij3=j=1ni=jnj3(change the order of summation)=j=1n(n+1j)j3=(n+1)j=1nj3j=1nj4=(n+1)[n(n+1)2]2n(n+1)(2n+1)(3n2+3n1)30=n(n+1)(n+2)(3n2+6n+1)60, \begin{aligned} \sum_{i=1}^n \left[ \frac { n(n+1)}{2} \right ]^2 & = \sum_{i=1}^n \sum_{j=1}^i j^3 \\ & = \sum_{j=1}^n \sum_{i=j}^{n} j^3 \quad \mbox{(change the order of summation)} \\ & = \sum_{j=1}^n (n+1-j) j^3 \\ & = (n+1) \sum_{j=1}^n j^3 - \sum_{j=1}^n j^4 \\ & = (n+1) \left[ \frac { n(n+1)}{2} \right ]^2 - \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30} \\ & = \frac{ n(n+1) (n+2) (3n^2 + 6n + 1) } { 60 }, \\ \end{aligned}

which agrees with Aditya's (just more factored in the last step).

I saved a lot of ugly expansion / calculation by being smart with the summation, and trying to keep it simple instead of expanding it all out immediately. The change of summation is also a very useful trick, that is used in numerous other places (e.g. interchanging the order of integration via Fubini's theorem). You just have to be careful of the limits.

Note: You should also know what i \sum i and i2 \sum i^2 are equal to.

Calvin Lin Staff - 6 years, 5 months ago

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Got that. I won't ask you how to get the value for Σ\Sigman4n^{4}. I just couldn't follow how did you use change of summation ? I never knew about it. Calvin please explain the steps starting changing order of summation.

Nishant Sharma - 6 years, 5 months ago

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Do you know how to get the equation for n\sum {n} , n2\sum {n^2} and for n3\sum{n^3}? You can use the same way to derive the value of n4\sum {n^4}

Aditya Parson - 6 years, 5 months ago

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I think in the starting of the solution it should be summation i=1 to n of i(i+1)/2... Also, pls explain the change of summation rule and the steps following it.

Krishna Jha - 6 years, 5 months ago

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Can the order of summation be changed always??

A Brilliant Member - 6 years, 5 months ago

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It is not hard to guess that the answer is a polynomial in nn if you recall the formula for the sum of the first nn numbers and for the sum of the first nn squares, both of which are well-known. It suffices to find the degree of the polynomial and then compute the first few sums to solve for the coefficients. It is not particularly hard to just start trying different degrees and find the answer. Degree 1 clearly doesn't work, and trying small degrees you eventually find that degree 55 works.

If you want to be a tad cleverer, you might conjecture based on the well-known formulas mentioned above, especially if you also know the one for sum of cubes, that the sum of the first nnth powers is a degree n+1n+1 polynomial. Since the triangular numbers are given by a degree 22 polynomial, and we are taking squares of them, the terms grow roughly at the rate of k4k^4. This leads us to immediately suspect that a degree 55 polynomial is the answer, and we can fit coefficients as above.

There is also a more conceptual approach to this problem, motivated by basic calculus. Recall that integration is analogous to summation and that much of what we know about summation carries over to integration (linearity, etc). In fact, the analogy also goes the other way: we can use what we know about integration to compute this sum!

What is the standard method of solving integrals? If we want the definite integral of g(x)g(x) over some interval, we try to find a function f(x)f(x) so that f(x)=g(x)f'(x)=g(x), then use the fundamental theorem of calculus to find the value of the integral by computing ff at the endpoints and subtracting the resulting values. A similar approach works here. I will illustrate it with a simpler, related problem that we already know how to solve with other methods: summing the first nn integers.

Let f(k)f(k) be a function on the integers, and define the 'discrete derivative' of f(k)f(k) to be f(k)=f(k+1)f(k)f'(k)=f(k+1)-f(k). Suppose we want to find the sum 1ng(k)\sum_1^n g(k), where g(k)g(k) is a function on the integers. If we can find hh so that h(k)=g(k)h'(k)=g(k), then we have

1ng(k)=1nf(k+1)f(k)=f(n+1)f(1).\displaystyle \sum_1^n g(k) = \sum_1^n f(k+1)-f(k) = f(n+1)-f(1).

Note how this is completely analogous to evaluating the antiderivative of an integral at its endpoints.

Returning to our problem, we want to find ff so that f(k)=kf'(k)=k. Motivated by standard calculus, we might think to try a degree 22 polynomial. This is also easily guessable. We can again solve for the coefficients in f(k+1)f(k)=kf(k+1)-f(k)=k, and find that f(k)=12(k2+k)+Cf(k)= \frac{1}{2}(k^2+k)+C, where CC is any constant. Then f(n+1)f(1)f(n+1)-f(1) is n(n+1)2\frac{n(n+1)}{2}, and we recover the standard formula.

This method is known as the 'calculus of finite differences,' and it generalizes in the obvious way to give a solution to the original problem. It also provides a standard method for approaching these kinds of problems.

Patrick Brown - 6 years, 5 months ago

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Sn=n(n+1)(6n3+24n2+26n+4)120S_n=\frac{n(n+1)(6n^3+24n^2+26n+4)}{120}

Aditya Parson - 6 years, 5 months ago

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Hey, what I eagerly wanted was the procedure.

Nishant Sharma - 6 years, 5 months ago

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Yeah sure. First develop a function that gives this sequence. Then just use summation. That is all.

Aditya Parson - 6 years, 5 months ago

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@Aditya Parson Seems that you don't wan't to reveal your method of solving(not in an offensive mood). I am not getting a series for which it is easy to find the sum. Rather i am getting terms like n4n^{4} in the general term, the sum to n terms of which is not known to me.

Nishant Sharma - 6 years, 5 months ago

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@Nishant Sharma n4=n(n+1)(2n+1)(3n2+3n1)30\sum n^4=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}

That is all you need to know.

Aditya Parson - 6 years, 5 months ago

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@Nishant Sharma hey, first of all try to get a sum involving n^4, it's a easy problem just huge calculations

Pratik Singhal - 6 years, 5 months ago

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how...?

A Former Brilliant Member - 6 years, 5 months ago

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Huge calculations. If I write it here using latex, it will take half an hour to say the least.

Aditya Parson - 6 years, 5 months ago

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@Aditya Parson kk.....do you have a worked out solution?

A Former Brilliant Member - 6 years, 5 months ago

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@A Former Brilliant Member Yeah, but I will need some time to make it look representable and understandable.

Aditya Parson - 6 years, 5 months ago

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@Aditya Parson k...then leave it....if u have time then only....post it...:|

A Former Brilliant Member - 6 years, 5 months ago

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@A Former Brilliant Member If you can wait for sometime, I can send you my workings via email.

Aditya Parson - 6 years, 5 months ago

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@Aditya Parson kk.....write ur e-mail...id here i m enndin a blank mail on it

A Former Brilliant Member - 6 years, 5 months ago

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but i got this solution as sum=(n)(n+1)(n+2)*(3n^2+6n+1)/15

Sayan Chaudhuri - 6 years, 5 months ago

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It is wrong. Put n=1 I get 4.

Aditya Parson - 6 years, 5 months ago

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   S(n)  =  1^2 + 3^2 + 6^2 + 10^2 + . . . . . . . . . . . . . . . . . . .+T(n).-----------(1)
- S(n) = 0 + 1^2 + 3^2 + 6^2 + 10^2 + . . . . . . . . . . . . . +T(n-1) + T(n) -----------(2) 0 = 1 + 8 + 27 + 64 + 125 + . . . . . . . . . . . . . . +n^3 –T(n) ( subtract 1 from 2 ) T(n) = sigma n^3 Therefore sigma T(n) = sigma of sigma n^3 S(n)= sum of series S(n) = sigma T(n) = sigma of sigma n^3 T(n)=n th term

Saichandan Bussireddy - 6 years, 5 months ago

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you could use newtons interpolation formula

Kishore Saldanha - 6 years, 5 months ago

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http://www.wolframalpha.com/widget/widgetPopup.jsp?p=v&id=d983db47634e1936eb568b79884012ac&title=Summation%20Calculator&theme=blue&i0=i%5E2&i1=10&i2=11&podSelect=&includepodid=Result&showAssumptions=1&showWarnings=1

Djordje Marjanovic - 6 years, 5 months ago

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