Sum the series: \(1^{2}\) + \(3^{2}\) + \(6^{2}\) + \(10^{2}\) + \(15^{2}\) + ..........n terms

Please give me the shortest way to solve this problem.

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## Comments

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

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

With this, we can write the sum as

\[ \begin{align} \sum_{i=1}^n \left[ \frac { n(n+1)}{2} \right ]^2 & = \sum_{i=1}^n \sum_{j=1}^i j^3 \\ & = \sum_{j=1}^n \sum_{i=j}^{n} j^3 \quad \mbox{(change the order of summation)} \\ & = \sum_{j=1}^n (n+1-j) j^3 \\ & = (n+1) \sum_{j=1}^n j^3 - \sum_{j=1}^n j^4 \\ & = (n+1) \left[ \frac { n(n+1)}{2} \right ]^2 - \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30} \\ & = \frac{ n(n+1) (n+2) (3n^2 + 6n + 1) } { 60 }, \\ \end{align} \]

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 \( \sum i \) and \( \sum i^2 \) are equal to.

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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.

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Got that. I won't ask you how to get the value for \(\Sigma\)\(n^{4}\). 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.

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

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Can the order of summation be changed always??

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It is not hard to guess that the answer is a polynomial in \(n\) if you recall the formula for the sum of the first \(n\) numbers and for the sum of the first \(n\) 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 \(5\) 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 \(n\)th powers is a degree \(n+1\) polynomial. Since the triangular numbers are given by a degree \(2\) polynomial, and we are taking squares of them, the terms grow roughly at the rate of \(k^4\). This leads us to immediately suspect that a degree \(5\) 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 \(g(x)\) over some interval, we try to find a function \(f(x)\) so that \(f'(x)=g(x)\), then use the fundamental theorem of calculus to find the value of the integral by computing \(f\) 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 \(n\) integers.

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

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

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

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

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

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but i got this solution as sum=(n)

(n+1)(n+2)*(3n^2+6n+1)/15Log in to reply

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

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how...?

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

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Comment deleted May 10, 2013

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thanksss alot :)

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Hey, what I eagerly wanted was the procedure.

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

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That is all you need to know.

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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

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you could use newtons interpolation formula

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