Question

Sol. [Key Concept: Pair consecutive cubes using $n^{3} - (n{+ 1)}^{3} = - 3n^{2} - 3n - 1$ for $n = 1,3,dots,13$, sum the seven pairs, then add the last term $15^{3}$.]

$$\left( 1^{3} - 2^{3} \right) + \left( 3^{3} - 4^{3} \right) + \ldots{.13}^{3} - 14^{3} + 15^{3} $$use $n^{3} - (n + 1)^{3} = - 3n^{2} - 3n - 1$

put $n = 1,3,5,\ldots.13$ ( 7 terms)

sum of pairs $= - 1519$

add last term $= 15^{3}$

$$\Rightarrow 3375 - 1519 = 1856 $$