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# Sequences Problem

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.

Note by Nishant Sharma
4 years, 10 months ago

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As 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.

Staff - 4 years, 10 months ago

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.

- 4 years, 10 months ago

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.

- 4 years, 10 months ago

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 $$\sum {n^4}$$

- 4 years, 10 months ago

Can the order of summation be changed always??

- 4 years, 10 months ago

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.

- 4 years, 10 months ago

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

- 4 years, 10 months ago

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

- 4 years, 10 months ago

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

- 4 years, 10 months ago

how...?

- 4 years, 10 months ago

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

- 4 years, 10 months ago

kk.....do you have a worked out solution?

- 4 years, 10 months ago

Yeah, but I will need some time to make it look representable and understandable.

- 4 years, 10 months ago

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

- 4 years, 10 months ago

If you can wait for sometime, I can send you my workings via email.

- 4 years, 10 months ago

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

- 4 years, 10 months ago

Comment deleted May 10, 2013

mail sent...

- 4 years, 10 months ago

Please hold on for a few minutes.

- 4 years, 10 months ago

would u please send the solution to me also....?...somehow i got wrong solution....i would be gratified if you kindly send it to ....metronetizen@gmail.com.....thanking you.....

- 4 years, 10 months ago

Sent.

- 4 years, 10 months ago

got it...thanxx a lot...

- 4 years, 10 months ago

Sure brother :D

- 4 years, 10 months ago

hmmm...:)

- 4 years, 10 months ago

Sent.

- 4 years, 10 months ago

got it...:)
thanksss alot :)

- 4 years, 10 months ago

please send me the solution too! id is codymartin7@rediffmail.com

- 4 years, 10 months ago

Hey, what I eagerly wanted was the procedure.

- 4 years, 10 months ago

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

- 4 years, 10 months ago

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 $$n^{4}$$ in the general term, the sum to n terms of which is not known to me.

- 4 years, 10 months ago

$$\sum n^4=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$$

That is all you need to know.

- 4 years, 10 months ago

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

- 4 years, 10 months ago

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

- 4 years, 10 months ago

you could use newtons interpolation formula

- 4 years, 10 months ago

   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

- 4 years, 10 months ago