Square Root of Sum of Sequential Cubes Equal to Sum of Sequence

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——————– Derivation for Sum of first N Squares ———————-

N^3 - (N-1)^3 = N^3 - (N^3 - 3N^2 + 3N -1) = 3N^2 - 3N + 1

(N-0)^3 - (N-1)^3 = 3(N-0)^2 - 3(N+1)+1

(N-1)^3 - (N-2)^3 = 3(N-1)^2 - 3( N-1)+1

(N-2)^3 - (N-1)^3 = 3(N-2)^2 - 3( N-2)+1

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

(1)^3 - (0)^3 = 3(0)^2 - 3(0)+1

—————————–Summing——————————————-

$N^3 = 3 sum {1}{N}{N^2} - 3 sum {1}{N}{N} + N$

$N^3 = 3 sum {1}{N}{N^2} - (3N^2+3N)/2 + N$

$N^3 = 3 sum {1}{N}{N^2} - (3N^2+N)/2$

$N^3 + (3N^2+N)/2 = 3 sum {1}{N}{N^2}$

$3 sum {1}{N}{N^2} = N^3 + (3N^2+N)/2$

$3 sum {1}{N}{N^2} = ( 2N^3 + 3N^2+N )/2$

$sum {1}{N}{N^2} = N( 2N^2 + 3N+1 )/6$

$sum {1}{N}{N^2} = {N(2N+1)(N+1)}/6$