An algebra problem by Anshuman Shreshth

What is given to us is:

$\sum_{i=1}^K i^3 - \sum_{i=1}^K i = 1980$

We will use two identities: $\displaystyle \sum_{i=1}^n i^3 = \left(\dfrac{n(n+1)}{2}\right)^2\quad \textrm{and}\quad \displaystyle \sum_{i=1}^n i = \dfrac{n(n+1)}{2}$

Let $X=\dfrac{K(K+1)}{2}$ . The given sum can now be written as:

$X^2-X-1980=0 \implies (X-45)(X+44)=1980 \implies X=45$

The negative solution is rejected since $K$ cannot be negative and hence $X$ also cannot be negative. Similarly, we'll reject the negative $K$ in the next step. Now,

$X=\dfrac{K(K+1)}{2}=45\\ \implies K^2+K-90=0 \implies (K-9)(K+10)=0 \implies \boxed{K=9}$

Use the formula: ${ 1 }^{ 3 }+{ 2 }^{ 3 }+{ 3 }^{ 3 }+.....+{ n }^{ 3 }={ (\frac { n(n+1) }{ 2 } ) }^{ 2 }$ and $+2+3+.....+n=\frac { n(n+1) }{ 2 }$

As the problem follows: ${ (\frac { n(n+1) }{ 2 } ) }^{ 2 }-\frac { n(n+1) }{ 2 } =1980$

Combining the fraction and simplifying: ${ n }^{ 4 }+2{ n }^{ 3 }-{ n }^{ 2 }-n-7920=0$

Graph it and the approximate positive root is $9.997$ which is rounded to $\boxed{10}$

anshuman shreshth, the conclusion which you reached in the class is already a formulae.
that is the sum of cube of terms of an a.p. is equal to square of the sum of the terms of the a.p. .