JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Binomial Theorem

The given sum is a geometric series with first term $a=1$, common ratio $r=3$, and $n=2022$ terms. Using the formula for the sum of a geometric series:



Now, write $9=10-1$ and apply the binomial theorem:



All terms with $k\ge 2$ are divisible by 100. The relevant terms are:

• $k=0$: $(-1{)}^{1011}=-1$
• $k=1$: $\left(\genfrac{}{}{0ex}{}{1011}{1}\right)\cdot 10\cdot (-1{)}^{1010}=10110$

So $9^{1011}=100\lambda +10110-1=100\lambda +10109$ for some integer $\lambda$.

Therefore:


Since $5054=50\times 101+4$, we have $S=50(\lambda +101)+4$.

The remainder when divided by 50 is 4.