Question

the value of (1³+2³+3³+.....15³)-(1+2+3+.....15)​

Answer 1

Answer:

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Answer 2

We need to recall the following formulas.

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

Given:

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

$=[\frac{15(15+1)}{2} ]^2-\frac{15(15+1)}{2}$

$=[\frac{15(16)}{2} ]^2-\frac{15(16)}{2}$

$=[120 ]^2-120$

$=14400-120$

$=14280$

Hence, $(1^3+2^3+3^3+....+15^3)-(1+2+3+....+15)=14280$