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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 $$ = \frac{{{{\left( {n\left( {n + 1} \right)} \right)}^2}}}{4}$$
Average
$$ = \frac{{{{\left( {n\left( {n + 1} \right)} \right)}^2}}}{4}$$
$$\eqalign{ & \Rightarrow n = 49 \cr & \Rightarrow \frac{{49{{\left( {50} \right)}^2}}}{4} \cr & \Rightarrow 30625 \cr} $$
Average
$$ = \frac{{{{\left( {n\left( {n + 1} \right)} \right)}^2}}}{4}$$
$$\eqalign{ & \Rightarrow n = 49 \cr & \Rightarrow \frac{{49{{\left( {50} \right)}^2}}}{4} \cr & \Rightarrow 30625 \cr} $$
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