Alternating cubes

$\large = { 1 }^{ 3 }-{ 2 }^{ 3 }+{ 3 }^{ 3 }-{ 4 }^{ 3 }+\cdots+{ 99 }^{ 3 }-{ 100 }^{ 3 }+{ 101 }^{ 3 }$ $\large = ({ 1 }^{ 3 }+{ 2 }^{ 3 }+ { 3 }^{ 3 } + { 4 }^{ 3 }+\cdots+{ 100 }^{ 3 }) -2(2^3 + 4^3 + \cdots 100^3) + 101^3$ $\large = \left( \dfrac{100 \times 101}{2}\right)^2 - 2^4(1^3 + 2^3 + \cdots + 50^3) + 101^3$ $\large = 5050^2 - 16\left (\dfrac{50 \times 51}{2}\right)^2 + 101^3$ $\large = \boxed{522801}$

$\color{#3D99F6}{\text{Formual used}} : \large 1^3 + 2^3 + 3^3 + \cdots + n^3 = \left( \dfrac{n (n+1)}{2}\right)^2$

$\displaystyle \sum^{50}_{n=0}(2n+1)^3-\displaystyle \sum^{50}_{n=1}(2n)^3 \\ 1^3+\displaystyle \sum^{50}_{n=1}(2n+1)^3-\displaystyle \sum^{50}_{n=1}(2n)^3 \\ 1+\displaystyle \sum^{50}_{n=1}(8n^3+12n^2+6n+1)-\displaystyle \sum^{50}_{n=1}(8n^3) \\ 1+\cancel{8\displaystyle \sum^{50}_{n=1}n^3}+12 \displaystyle \sum^{50}_{n=1}n^2+6 \displaystyle \sum^{50}_{n=1}n+ \displaystyle \sum^{50}_{n=1}1-\cancel{8 \displaystyle \sum^{50}_{n=1}n^3} \\ 1+12\cdot \frac{50×51×101}{6}+6\cdot \frac{50×51}{2}+50 \\ \boxed{522801}$

${ 1 }^{ 3 }-{ 2 }^{ 3 }+{ 3 }^{ 3 }-{ 4 }^{ 3 }+\cdots+{ 99 }^{ 3 }-{ 100 }^{ 3 }+{ 101 }^{ 3 }$

$= { 1 }^{ 3 }+({ 3 }^{ 3 }-{ 2 }^{ 3 })+({ 5 }^{ 3 }-{ 4 }^{ 3 })+\cdots+({101 }^{ 3 }-{ 100 }^{ 3 })$

$= 1+(3-2)({ 3 }^{ 2 }+{ 2 }^{ 2 }+3 \times 2)+(5-4)({ 5 }^{ 2 }+{ 4 }^{ 2 }+5 \times 4)+\cdots+(101-100)({ 101 }^{ 2 }+{ 100 }^{ 2 }+101 \times 100)$

$= 1+({ 2 }^{ 2 }+{ 3 }^{ 2 }+ \cdots + { 100 }^{ 2 }+{ 101 }^{ 2 }) + (2 \times 3 + 4 \times 5 + \cdots + 100 \times 101)$

$= 1+ \sum _{ r=2 }^{ 101 }{ { r }^{ 2 } } + \sum _{ r=1 }^{ 50 }{ 2r\left( 2r+1 \right) }$

$= \sum _{ r=1 }^{ 101 }{ { r }^{ 2 } } + 4\sum _{ r=1 }^{ 50 }{ { r }^{ 2 } } + 2\sum _{ r=1 }^{ 50 }{ r }$

$= \cfrac { 1 }{ 6 } \left( 101 \right) \left( 102 \right) \left( 203 \right) + 4\left[ \cfrac { 1 }{ 6 } \left( 50 \right) \left( 51 \right) \left( 101 \right) \right] + 2\left[ \cfrac { 1 }{ 2 } \left( 50 \right) \left( 51 \right) \right]$

$= \large \color{#20A900}{\boxed{522801}}$

Let the sum be $S$ , then:

$\begin{aligned} S & = 1^3 - 2^3 + 3^3-4^3+5^3-6^3...+99^3 - 100^3+101^3 \\ & = \sum_{n=1}^{50} \left[(2n-1)^3-(2n)^3\right] + 101^3 \\ & = \sum_{n=1}^{50} \left[8n^3-12n^2+6n -1-8n^3\right] + 101^3 \\ & = \sum_{n=1}^{50} \left[-12n^2+6n -1 \right] + 101^3 \\ & = -12 \sum_{n=1}^{50} n^2 + 6 \sum_{n=1}^{50}n - \sum_{n=1}^{50} 1 + 101^3 \\ & = - 12 \times \frac{50(51)(101)}{6} + 6 \times \frac{50(51)}{2} - 50 + 101^3 \\ & = -515100+7650-50+1030301 \\ & = \boxed{522801} \end{aligned}$