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My teacher wrote the following identity out of the blue for finding out the sum of squares of first n natural numbers:

$r^3 - (r - 1)^3 = 3r^2 - 3r + 1$

Then, the value of r was substituted for 1, 2, 3, 4,..., n respectively.

There were n number of identities as follows:

$1^3-0^3=3*1^2-3*1+1$

$2^3-1^3=3*2^2-3*2+1$

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$n^3-(n-1)^3=3n^2-3n+1$

All of the identities were added to obtain

$n^3=3(1^2+2^2+3^2+. . . +n^2)-3(1+2+3+...+n)+n$

=>$n^3=3S$n$-3n(n+1)/{2}+n$

And lastly, the above identity was solved for Sn

My question: Why was the first identity in r chosen? Are there other identities which can be chosen?

S.Nep
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1 Answers1

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The first identity has been chosen to use telescoping for the LHS indeed

$$r^3 - (r - 1)^3 = 3r^2 - 3r + 1 \implies \sum (r^3 - (r - 1)^3) = \sum (3r^2 - 3r + 1)$$

$$\implies n^3=3S_n-3\sum r+n$$

and then find the required $S_n$ from the well known formula for $\sum r$.

In the same way we can find the sum of the cubes starting from

$$r^4 - (r - 1)^4 = 4r^3-6r^2+4r-1$$

For other methods to find the result refer to the related

user
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