Are you Smart in Simple Algebra #2?

Algebra Level 3

$\large{\sum_{k=1}^{n} \left(\dfrac{3k+1}{k+1} \cdot {2k \choose k} \right) = 68}$

Find the sum of all natural numbers $n$ such that it satisfies the above equation.

1 solution

Shauryam Akhoury
Sep 13, 2018

Since no term is negative,once the value reaches $68$ it never will and just keep on increasing with increasing $n$

When $n = 4$ ,

${2n \choose n}>68$

So for the sum being 68, $n<4$

And when $n=3$ the sum does equal $68$ and will never agian

Since there is just one value for $n$ the sum is $3$