JEE Main 2021 (Online) 26th February Morning ShiftQuestion 1
Let $m,n\in\mathbf{N}$ and $\gcd(2,n)=1$. If $30\binom{30}{0}+29\binom{30}{1}+\ldots+2\binom{30}{28}+1\binom{30}{29}=n\cdot2^{m}$, then $n+m$ is equal to $\_\_\_\_\_\_$ . (Here $\binom{n}{k}={}^{n}C_k$)
Binomial Theorem · Properties and Sums of Binomial Coefficients
Answer$45$
Solution
Concept. The sum is $\sum_{k=0}^{29}(30-k)\,{}^{30}C_k$; split it into two standard sums. Formulas used. (1) $\sum_{k=0}^{n}{}^{n}C_k=2^n$ and $\sum_{k=0}^{n}k\,{}^{n}C_k=n\,2^{n-1}$Step 1 - Split. $$\sum_{k=0}^{29}(30-k)\,{}^{30}C_k=30\sum_{k=0}^{29}{}^{30}C_k-\sum_{k=0}^{29}k\,{}^{30}C_k.$$ Step 2 - Each part. The $k=30$ term is missing. By (1): $$\sum_{k=0}^{29}{}^{30}C_k=2^{30}-1,\qquad\sum_{k=0}^{29}k\,{}^{30}C_k=30\cdot2^{29}-30.$$ Step 3 - Combine. $$30\left(2^{30}-1\right)-30\cdot2^{29}+30=30\cdot2^{29}=15\cdot2^{30}.$$ So $n=15$ (odd) and $m=30$. Step 4 - Value. $n+m=45$. Answer: 45