Question 62

The value of

$$(1^3 + 2^3 + 3^3 + ........ + 15^3) - (1 + 2 + 3 + ......... + 15)$$ is

Solution

Sum of $$n$$ consecutive natural number cubes = $$[\frac{n(n+1)}{2}]^2$$

and sum of $$n$$ consecutive natural numbers = $$\frac{n(n+1)}{2}$$

Expression : $$(1^3 + 2^3 + 3^3 + ........ + 15^3) - (1 + 2 + 3 + ......... + 15)$$

= $$(\frac{15\times16}{2})^2-(\frac{15\times16}{2})$$

= $$(120)^2-120$$

= $$14400-120=14280$$

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