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Find the average of cubes of first 49 positive integers.
Answer & Solution
Correct Answer:
Option
A
Sum of cubes of first $$n$$ positive consecutive numbers is
$$\eqalign{ & = \frac{{{{\left\{ {n\left( {n + 1} \right)} \right\}}^2}}}{4} \cr & \therefore \,{\text{Average}} = \frac{{n{{\left( {n + 1} \right)}^2}}}{4} \cr & \Rightarrow n = 49 \cr & \Rightarrow \frac{{49{{\left( {50} \right)}^2}}}{4} = 30625 \cr} $$
$$\eqalign{ & = \frac{{{{\left\{ {n\left( {n + 1} \right)} \right\}}^2}}}{4} \cr & \therefore \,{\text{Average}} = \frac{{n{{\left( {n + 1} \right)}^2}}}{4} \cr & \Rightarrow n = 49 \cr & \Rightarrow \frac{{49{{\left( {50} \right)}^2}}}{4} = 30625 \cr} $$
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