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I tried with $$(2n-1)^3-(2n-2)^3=12n^2-18n+7$$ Now, forming partial sums for $n=1,2,...$ $$(1^3-0^3)+(3^3-2^3)+...+((2n-1)^3-(2n-2)^3)=12(1^2+...+n^2)-18(1+...+n)+7n$$ How to express $(1^3-0^3)+(3^3-2^3)+...+((2n-1)^3-(2n-2)^3)$ in general term?

Could some other method be used?

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Hint:

Your sum is

$$\sum_{k=1}^n (2k-1)^2=\sum_{k=1}^n 4k^2-\sum_{k=1}^n 4k+\sum_{k=1}^n1=4\sum_{k=1}^n k^2-4\sum_{k=1}^n k+n$$

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HINT:

Let $f(n)=a_0+a_1n+a_2n^2+a_3n^3$

Write $(2n-1)^2=f(n)-f(n-1)$ to find $a_0,a_1,a_2,a_3$

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why not use the formula for $\Sigma r^2$?

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  • $\begingroup$ When I use it directly, $S_{n^2}=\frac{n(n+1)(2n+1)}{6}$ and apply it for $S_{(2n-1)^2}$ I get $S_{(2n-1)^2}=\frac{n(2n-1)(4n-1)}{3}$ I have checked for $n=3$ and this is not correct. $\endgroup$ – user300045 Jun 16 '15 at 10:29
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    $\begingroup$ @no_name: You need to substract the even terms. $\endgroup$ – servabat Jun 16 '15 at 10:36
  • $\begingroup$ @no_name You can use it in user's "calculus" hint (you'll also need to use $1+2+\cdots+n=\frac{n(n+1)}{2}$ though). Also as servabat suggests $S_{(2n-1)^2}$ is $1^2+2^2+3^2+\cdots+(2n-1)^2$ and not $1^2+3^2+\cdots+(2n-1)^2$. You need to subtract the even terms inbetween. $\endgroup$ – user26486 Jun 16 '15 at 10:36
  • $\begingroup$ See Ihf's answer for how to subtract the even terms. $\endgroup$ – user26486 Jun 16 '15 at 10:50
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Note that $$k^2=\binom k2 +\binom {k+1}2$$ Hence $$\begin{align} \sum_{i=1}^n(2i-1)^2&=\sum_{i=1}^n {2i-1\choose 2}+{2i\choose 2}\\ &=\sum_{i=1}^{2n}\binom i2\\ &={2n+1\choose 3}\qquad\blacksquare\\ \end{align}$$

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I love Telescoping magic $$\dfrac{4n^3}3-\dfrac{4(n-1)^3}{3}=??$$

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

$\displaystyle \quad= \sum_{k=1}^{2n} k^2 - \sum_{k=1}^{n} (2k)^2 $

$\displaystyle \quad= \sum_{k=1}^{2n} k^2 - 4\sum_{k=1}^{n} k^2 $

$ \quad= \dfrac{(2n)(2n+1)(4n+1)}{6}-\dfrac{4n(n+1)(2n+1)}{6} $

$ \quad= \dfrac{n (2 n-1) (2 n+1)}{3} $

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