JEE Main 2025 — 4 April, Evening Shift. Previous Year Question.
Problem
If $1^2\cdot{}^{15}C_1+2^2\cdot{}^{15}C_2+3^2\cdot{}^{15}C_3+\cdots+15^2\cdot{}^{15}C_{15}$ is equal to $2^n3^m5^k$, then $m+n+k$ is equal to?
Key insight. A sum like $\sum r^2\binom{n}{r}$ isn’t something to compute term by term — it has a clean closed form, $n(n+1)2^{n-2}$, derived by writing $r^2=r(r-1)+r$ and using the standard identities for $\sum r\binom{n}{r}$ and $\sum r(r-1)\binom{n}{r}$. Once the sum collapses to a single product, prime-factorising it is straightforward.
Watch this explained step by step → · More on the Shiwam’s Classes channel
Approach
Use the standard identity $\displaystyle\sum_{r=0}^{n} r^2\binom{n}{r} = n(n+1)2^{n-2}$ (valid for $n\geq2$), applying it with $n=15$. Compute the resulting product, then factorise it completely into powers of $2$, $3$, and $5$ to read off $n$, $m$, $k$ (matching the problem’s own labelling, which reuses $n$ for the power of $2$).
Solution
Step 1 — Apply the standard identity
$$\sum_{r=0}^{15} r^2\binom{15}{r} = 15\times16\times2^{15-2} = 15\times16\times2^{13}$$
(The $r=0$ term contributes $0$, so this matches the sum starting from $r=1$.)
Step 2 — Simplify the numeric factors
$$15\times16 = 240$$
Step 3 — Factorise 240
$$240 = 16\times15 = 2^4\times3\times5$$
Step 4 — Combine with the remaining power of 2
$$\sum r^2\binom{15}{r} = 240\times2^{13} = 2^4\times3\times5\times2^{13} = 2^{17}\times3^1\times5^1$$
Step 5 — Match to the given form 2ⁿ3ᵐ5ᵏ
$$n=17, \qquad m=1, \qquad k=1$$
Step 6 — Compute m + n + k
$$m+n+k = 1+17+1 = 19$$
Answer
$$19$$
Common mistakes
- Trying to compute all 15 terms individually and sum them. This is technically possible but extremely tedious; the closed-form identity $n(n+1)2^{n-2}$ turns a 15-term sum into a two-second calculation.
- Forgetting to fully factorise $15\times16$ before combining with $2^{13}$. Leaving the answer as $240\times2^{13}$ without breaking $240$ down into $2^4\times3\times5$ makes it impossible to correctly read off the individual exponents $n$, $m$, $k$.
Practise next
- Find $\displaystyle\sum_{r=0}^{8} r^2\binom{8}{r}$, write it in the form $2^a3^b$, and compute $a+b$, using the same identity.
Show answer
$4608 = 2^9\cdot3^2$, so $a+b=11$. The identity is $\sum_{r=0}^{n} r^2\binom{n}{r}=n(n+1)2^{n-2}$.
With $n=8$ that is $8\cdot9\cdot2^{6}=72\times64=4608$, and $4608=512\times9=2^{9}\cdot3^{2}$. So $a=9$, $b=2$ and $a+b=11$.
The identity comes from $r\binom{n}{r}=n\binom{n-1}{r-1}$ applied twice, which is the same absorption step the problem above turns on.
Doubts are answered by Shiwam, usually within a day. Ask about this question specifically — a general question about the chapter is better asked in class.