JEE Main 2025 Apr 2 Shift 2, Mathematics Q15: Properties Of Binomial Coefficient

Rewrite the summand: $\frac{10^{r+1}-1}{10^r}=10-10^{-r}$. So

$S=\sum _{r=0}^{10}(10-10^{-r}){}^{11}{C}_{r+1}=10\sum _{r=0}^{10}{}^{11}{C}_{r+1}-\sum _{r=0}^{10}10^{-r}{}^{11}{C}_{r+1}$.

The first sum, with index shift $k=r+1$, runs $k=1$ to $11$: ${\sum }_{k=1}^{11}{}^{11}{C}_k=2^{11}-1$.

For the second, factor a $10^{-1}$ inside and use ${\sum }_{k=1}^{11}{}^{11}{C}_{k}10^{-k}=(1+10^{-1}{)}^{11}-1$:

${\sum }_{r=0}^{10}10^{-r}{}^{11}{C}_{r+1}=10{\sum }_{k=1}^{11}{}^{11}{C}_{k}10^{-k}=10\left[{\left(\frac{11}{10}\right)}^{11}-1\right]$.

So $S=10(2^{11}-1)-10\left[\frac{11^{11}}{10^{11}}-1\right]=10\cdot 2^{11}-\frac{11^{11}}{10^{10}}=\frac{20^{11}-11^{11}}{10^{10}}$.

Matching with $\frac{{\alpha }^{11}-11^{11}}{10^{10}}$ gives $\alpha =20$.