Question

If $1^3+2^3+3^3+....+k^3=8281$, then find $1+2+3+....+k$

Answer 1

In the given series $1^3+2^3+3^3++....+k^3=8281$, $S_k=8281$ represents the sum of cubes of natural numbers upto $k$. From this we have to find sum of natural numbers upto $k$. Thus, we can write the formula instead of the series and that is:

Sum of cubes of $n$ natural numbers is $S_n={\left[\frac{n(n+1)}{2}\right]}^2$, therefore, we have

$S_k={\left[\frac{k(k+1)}{2}\right]}^2\Rightarrow 8281={\left[\frac{k(k+1)}{2}\right]}^2\Rightarrow \frac{k(k+1)}{2}=\sqrt{8281}\Rightarrow \frac{k(k+1)}{2}=\mathrm{91......}\left(1\right)$

We know that the sum of natural numbers is $S_n=\frac{n(n+1)}{2}$, thus consider the sum as follows:

$1+2+3+.......+k=S_k\Rightarrow 1+2+3+.......+k=\frac{k(k+1)}{2}\Rightarrow 1+2+3+.......+k=91\left(from\right(1\left)\right)$

Hence $1+2+3+.......+k=91$.