Find the sum of consecutive cubes Algebra II

885

775

572

654

15,500

11,100

12,200

14,400

975

860

0

1125

99,500

103,600

110,500

95,200

\dfrac{k_1^2\cdot\left(k_1+1\right)^2-400}{4}

\dfrac{2k_1^2\cdot\left(k_1+1\right)^2-200}{4}

\dfrac{3k_1^2\cdot\left(k_1+1\right)^2-100}{4}

\dfrac{4k_1^2\cdot\left(k_1+1\right)^2-500}{4}

\dfrac{5k_1^2\cdot\left(k_1+1\right)^2-200}{4}

\dfrac{3k_1^2\cdot\left(k_1+1\right)^2-100}{4}

\dfrac{k_1^2\cdot\left(k_1+1\right)^2-400}{4}

\dfrac{2k_1^2\cdot\left(k_1+1\right)^2-100}{4}

\dfrac{5k_2^3\left(k_2+1\right)}{6}-\dfrac{5k_1^3\left(k_1+1\right)}{6}

\dfrac{3k_2^3\left(k_2+1\right)}{6}-\dfrac{3k_1^3\left(k_1+1\right)}{6}

\dfrac{k_2^3\left(k_2+1\right)}{6}-\dfrac{k_1^3\left(k_1+1\right)}{6}

\dfrac{k_2^2\left(k_2+1\right)^2}{4}-\dfrac{k_1^2\left(k_1+1\right)^2}{4}