If Σr²(³⁰Cᵣ)²/(³⁰Cᵣ₋₁)=α·2²⁹ From r=1 to 30, Find α

JEE Main 2025 — 28 January, Morning Shift. Previous Year Question.

Problem

If $\displaystyle\sum_{r=1}^{30} \dfrac{r^2\,(^{30}C_r)^2}{^{30}C_{r-1}} = \alpha\cdot 2^{29}$, then find the value of $\alpha$.

Key insight. The ratio $\dfrac{{}^{30}C_r}{{}^{30}C_{r-1}}$ simplifies to a clean fraction in terms of $r$ using the factorial definition of binomial coefficients — and once that ratio is substituted in, the whole sum can be split into two pieces, each of which telescopes into a binomial expansion of $2^{29}$ or $2^{28}$.

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Approach

The expression mixes ${}^{30}C_r$ (squared) with ${}^{30}C_{r-1}$ in the denominator — a strong hint to expand both in factorial form and simplify their ratio before attempting the sum. Once the ratio collapses to something involving $(31-r)$ and a single power of ${}^{30}C_r$-like term, the $r^2$ out front can be split as $r\cdot r$, and each factor of $r$ can be absorbed into the binomial coefficient using the standard trick $r\cdot {}^nC_r = n\cdot {}^{n-1}C_{r-1}$, repeated twice.

Solution

Step 1 — Simplify the ratio $\dfrac{({}^{30}C_r)^2}{{}^{30}C_{r-1}}$

Writing everything in factorials:

$${}^{30}C_r = \frac{30!}{r!(30-r)!}, \qquad {}^{30}C_{r-1} = \frac{30!}{(r-1)!(31-r)!}$$

Dividing one power of ${}^{30}C_r$ by ${}^{30}C_{r-1}$:

$$\frac{{}^{30}C_r}{{}^{30}C_{r-1}} = \frac{(r-1)!(31-r)!}{r!(30-r)!} = \frac{31-r}{r}$$

So the original term becomes:

$$r^2\cdot\frac{({}^{30}C_r)^2}{{}^{30}C_{r-1}} = r^2\cdot {}^{30}C_r\cdot\frac{31-r}{r} = r(31-r)\,{}^{30}C_r$$

Step 2 — Split $r(31-r)$ into two sums

$$\sum_{r=1}^{30} r(31-r)\,{}^{30}C_r = 31\sum_{r=1}^{30} r\cdot{}^{30}C_r \;-\; \sum_{r=1}^{30} r^2\cdot{}^{30}C_r$$

Step 3 — Evaluate $\sum r\cdot{}^{30}C_r$

Using the identity $r\cdot{}^{30}C_r = 30\cdot{}^{29}C_{r-1}$:

$$\sum_{r=1}^{30} r\cdot{}^{30}C_r = 30\sum_{r=1}^{30}{}^{29}C_{r-1} = 30\sum_{j=0}^{29}{}^{29}C_j = 30\cdot 2^{29}$$

Step 4 — Evaluate $\sum r^2\cdot{}^{30}C_r$

This needs the same trick applied twice. Writing $r^2\cdot{}^{30}C_r = r\cdot(30\cdot{}^{29}C_{r-1})= 30\cdot r\cdot{}^{29}C_{r-1}$, and then substituting $r = (r-1)+1$:

$$30\sum_{r=1}^{30}\big[(r-1)+1\big]{}^{29}C_{r-1} = 30\left[\sum_{r=1}^{30}(r-1)\,{}^{29}C_{r-1} + \sum_{r=1}^{30}{}^{29}C_{r-1}\right]$$

The second piece is $\sum_{j=0}^{29}{}^{29}C_j = 2^{29}$. For the first piece, using $(r-1)\cdot{}^{29}C_{r-1} = 29\cdot{}^{28}C_{r-2}$ (the same trick again, shifted by one index):

$$\sum_{r=1}^{30}(r-1)\,{}^{29}C_{r-1} = 29\sum_{r=2}^{30}{}^{28}C_{r-2} = 29\sum_{j=0}^{28}{}^{28}C_j = 29\cdot 2^{28}$$

So:

$$\sum_{r=1}^{30} r^2\cdot{}^{30}C_r = 30\left[29\cdot 2^{28} + 2^{29}\right] = 30\cdot 2^{28}\big[29+2\big] = 30\cdot 31\cdot 2^{28}$$

Step 5 — Combine both pieces

$$31\sum r\cdot{}^{30}C_r – \sum r^2\cdot{}^{30}C_r = 31\cdot 30\cdot 2^{29} – 30\cdot 31\cdot 2^{28}$$

Factoring out $30\cdot 31\cdot 2^{28}$:

$$30\cdot 31\cdot 2^{28}\big(2-1\big) = 30\cdot 31\cdot 2^{28}$$

Converting to match the required $2^{29}$ form: $30\cdot 31\cdot 2^{28} = \dfrac{30\cdot 31}{2}\cdot 2^{29} = 465\cdot 2^{29}$.

Answer

Common mistakes

• Not simplifying $\dfrac{{}^{30}C_r}{{}^{30}C_{r-1}}$ first, and instead trying to work with $\left({}^{30}C_r\right)^2$ and ${}^{30}C_{r-1}$ as separate unrelated sums — the whole problem only becomes tractable once that ratio collapses to $\dfrac{31-r}{r}$.
• Applying $r\cdot{}^nC_r = n\cdot{}^{n-1}C_{r-1}$ only once when the $r^2$ term needs it applied twice, with careful index-shifting the second time — skipping this leaves a stray $r$ that can’t be summed directly.

Practise next

• Evaluate $\displaystyle\sum_{r=1}^{20} \dfrac{r(^{20}C_r)^2}{^{20}C_{r-1}}$ using the same ratio-simplification approach, adjusting for a single power of $r$ instead of $r^2$.