Question
Observe that $${1^3} = 1,{2^3} = 3 + 5,{3^3} = 7 + 9 + 11,{4^3} = 13 + 15 + 17 + 19.$$ Then $${n^3}$$ as a similar series is
A.
$$\left[ {2\left\{ {\frac{{n\left( {n - 1} \right)}}{2} + 1} \right\} - 1} \right] + \left[ {2\left\{ {\frac{{\left( {n + 1} \right)n}}{2} + 1} \right\} + 1} \right] + ..... + \left[ {2\left\{ {\frac{{\left( {n + 1} \right)n}}{2} + 1} \right\} + 2n - 3} \right]$$
B.
$$\left( {{n^2} + n + 1} \right) + \left( {{n^2} + n + 3} \right) + \left( {{n^2} + n + 5} \right) + ..... + \left( {{n^2} + 3n - 1} \right)$$
C.
$$\left( {{n^2} - n + 1} \right) + \left( {{n^2} - n + 3} \right) + \left( {{n^2} - n + 5} \right) + ..... + \left( {{n^2} + n - 1} \right)$$
D.
none of these
Answer :
$$\left( {{n^2} - n + 1} \right) + \left( {{n^2} - n + 3} \right) + \left( {{n^2} - n + 5} \right) + ..... + \left( {{n^2} + n - 1} \right)$$
Solution :
$${1^3} = 1 \cdot \left( {1 - 1} \right) + 1,{2^3} = \left( {2 \cdot 1 + 1} \right) + 5,{3^3} = \left( {3 \cdot 2 + 1} \right) + 9 + 11,{4^3} = \left( {4 \cdot 3 + 1} \right) + 15 + 17 + 19,{\text{e}}{\text{.t}}{\text{.c}}{\text{.}}$$
$$\therefore \,\,{n^3} = \left\{ {n \cdot \left( {n - 1} \right) + 1} \right\} + .....,$$ next term being 2 more than the previous
$$\therefore \,\,{n^3} = \left( {{n^2} - n + 1} \right) + \left( {{n^2} - n + 3} \right) + .....$$