Another method to solve the summation series from Quiz 9

a

The summation of series from quiz 9 (quiz-9-mock-imo-algebra-contest) can be tackled using another method.

First, note that $n^2-2n+1,\,n^2-1,\,n^2+2n+1$ all can be factored as:

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

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

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

Therefore, the given sum

[MATH]\sum_{n=4}^{999}\dfrac{1}{\sqrt[3]{n^2-2n+1}+\sqrt[3]{n^2+2n+1}+\sqrt[3]{n^2-1}}[/MATH]

is hence is equivalent to

[MATH]\sum_{n=4}^{999}\dfrac{1}{((n-1)^2)^\frac{1}{3}+((n+1)^2)^\frac{1}{3}+((n-1)(n+1))^\frac{1}{3}}[/MATH]

[MATH]=\sum_{n=4}^{999}\dfrac{1}{((n-1)^\frac{1}{3})^2+((n+1)^\frac{1}{3})^2+((n-1)(n+1))^\frac{1}{3}}[/MATH]

[MATH]=\color{black}\sum_{n=4}^{999}\dfrac{1}{(\color{yellow}\bbox[5px,purple]{(n-1)^\frac{1}{3}}\color{black})^2+(\color{yellow}\bbox[5px,green]{(n+1)^\frac{1}{3}}\color{black})^2+(\color{yellow}\bbox[5px,purple]{(n-1)^\frac{1}{3}}\color{yellow}\bbox[5px,green]{(n+1)^\frac{1}{3}}\color{black}\color{black})}[/MATH]

[MATH]=\color{black}\sum_{n=4}^{999}\dfrac{1}{(\color{yellow}\bbox[5px,purple]{a}\color{black})^2+(\color{yellow}\bbox[5px,green]{b}\color{black})^2+(\color{yellow}\bbox[5px,purple]{a}\color{yellow}\bbox[5px,green]{b}\color{black}\color{black})}[/MATH](*)

I will stop momentarily here, because I want you to focus on the last expression above, to see if something has (or will) come to you mind when you begin to think, reflect and recall.

I have faith in you that you are already hit on a great idea, haven't you?

Yeah! That exact expression [MATH](\color{yellow}\bbox[5px,purple]{a}\color{black})^2+(\color{yellow}\bbox[5px,green]{b}\color{black})^2+(\color{yellow}\bbox[5px,purple]{a}\color{yellow}\bbox[5px,green]{b}\color{black}\color{black})[/MATH] is the factor that we would get when we factor the difference of cubes:

[MATH]\color{black}\bbox[5px,orange]{a^3-b^3=(a-b)(a^2+ab+b^2)}[/MATH]

We need to make optimal use of this identity from now on to our problem, that is to say, we need to multiply the top and bottom of the expression in (*) with the quantity $a-b$, i.e. $(n-1)^{\frac{1}{3}}-(n+1)^{\frac{1}{3}}$ so to obtain:

[MATH]\sum_{n=4}^{999}\dfrac{1}{\sqrt[3]{n^2-2n+1}+\sqrt[3]{n^2+2n+1}+\sqrt[3]{n^2-1}}[/MATH]

[MATH]=\sum_{n=4}^{999}\dfrac{1}{((n-1)^\frac{1}{3})^2+((n+1)^\frac{1}{3})^2+((n-1)(n+1))^\frac{1}{3}}[/MATH]

[MATH]=\color{black}\sum_{n=4}^{999}\dfrac{1}{(\color{yellow}\bbox[5px,purple]{(n-1)^\frac{1}{3}}\color{black})^2+(\color{yellow}\bbox[5px,green]{(n+1)^\frac{1}{3}}\color{black})^2+(\color{yellow}\bbox[5px,purple]{(n-1)^\frac{1}{3}}\color{yellow}\bbox[5px,green]{(n+1)^\frac{1}{3}}\color{black}\color{black})}[/MATH]

[MATH]=\color{black}\sum_{n=4}^{999}\dfrac{1}{(\color{yellow}\bbox[5px,purple]{a}\color{black})^2+(\color{yellow}\bbox[5px,green]{b}\color{black})^2+(\color{yellow}\bbox[5px,purple]{a}\color{yellow}\bbox[5px,green]{b}\color{black}\color{black})}[/MATH](*)

[MATH]=\color{black}\sum_{n=4}^{999}\dfrac{1}{(\color{yellow}\bbox[5px,purple]{a}\color{black})^2+(\color{yellow}\bbox[5px,green]{b}\color{black})^2+(\color{yellow}\bbox[5px,purple]{a}\color{yellow}\bbox[5px,green]{b}\color{black}\color{black})}\dfrac{\cdot \color{black}(\color{yellow}\bbox[5px,purple]{a}\color{black}-\color{yellow}\bbox[5px,green]{b}\color{black})}{\cdot \color{black}(\color{yellow}\bbox[5px,purple]{a}\color{black}-\color{yellow}\bbox[5px,green]{b}\color{black})}[/MATH]

[MATH]=\sum_{n=4}^{999} \dfrac{\color{black}(\color{yellow}\bbox[5px,purple]{a}\color{black}-\color{yellow}\bbox[5px,green]{b}\color{black})}{\color{black}(\color{yellow}\bbox[5px,purple]{a^3}\color{black}-\color{yellow}\bbox[5px,green]{b^3}\color{black})}[/MATH]

[MATH]=\sum_{n=4}^{999}\dfrac{(n-1)^{\frac{1}{3}}-(n+1)^{\frac{1}{3}}}{((n-1)^{\frac{1}{3}})^3-((n+1)^{\frac{1}{3}})^3}[/MATH]

[MATH]=\sum_{n=4}^{999}\dfrac{(n-1)^{\frac{1}{3}}-(n+1)^{\frac{1}{3}}}{n-1-n-1}[/MATH]

[MATH]=\sum_{n=4}^{999}\dfrac{(n-1)^{\frac{1}{3}}-(n+1)^{\frac{1}{3}}}{-2}[/MATH]

[MATH]=\sum_{n=4}^{999}\dfrac{(n+1)^{\frac{1}{3}}-(n-1)^{\frac{1}{3}}}{2}[/MATH]

$=\dfrac{1}{2}[(5^{\frac{1}{3}}-3^{\frac{1}{3}})+(6^{\frac{1}{3}}-4^{\frac{1}{3}})+(7^{\frac{1}{3}}-5^{\frac{1}{3}})+(8^{\frac{1}{3}}-6^{\frac{1}{3}})+\cdots$


$\,\,\,\,\,\,\,\,\,\,\,\,+(998^{\frac{1}{3}}-996^{\frac{1}{3}})+(999^{\frac{1}{3}}-997^{\frac{1}{3}})+(1000^{\frac{1}{3}}-998^{\frac{1}{3}})]$

$=\dfrac{1}{2}[(\cancel{5^{\frac{1}{3}}}-3^{\frac{1}{3}})+(\cancel{6^{\frac{1}{3}}}-4^{\frac{1}{3}})+(\cancel{7^{\frac{1}{3}}}-\cancel{5^{\frac{1}{3}}})+(\cancel{8^{\frac{1}{3}}}-\cancel{6^{\frac{1}{3}}})+\cdots$


$\,\,\,\,\,\,\,\,\,\,\,\,+(\cancel{998^{\frac{1}{3}}}-\cancel{996^{\frac{1}{3}}})+(999^{\frac{1}{3}}-\cancel{997^{\frac{1}{3}}})+(1000^{\frac{1}{3}}-\cancel{998^{\frac{1}{3}}})]$

$=\dfrac{1}{2}[-3^{\frac{1}{3}}-4^{\frac{1}{3}}+999^{\frac{1}{3}}+1000^{\frac{1}{3}}]$

$=\dfrac{1}{2}[-3^{\frac{1}{3}}-4^{\frac{1}{3}}+999^{\frac{1}{3}}+10]$