If 20C1 + (2²) 20C2 + (3²) 20C3 + ....... + (20²) 20C20 = A(2^β), then the ordered pair (A, β) is eq

Q: Why do we multiply by $x$ after the first derivative? Why not just differentiate again immediately?

That is a brilliant question! If you differentiate immediately after the first step, you get $r(r-1)x^{r-2}$. That gives you $r^2 - r$, which is messy and doesn't isolate the $r^2$ term we need. Multiplying by $x$ resets the power of $x$ back to $x^r$. When you differentiate that, you get $r^2 x^{r-1}$. It is the secret 'reset button' that keeps the algebra clean and elegant!

Q: Is there a way to do this without calculus? It feels like cheating.

It is not cheating; it is using the most powerful tool in your arsenal! You could use combinatorial identities like $r \binom{n}{r} = n \binom{n-1}{r-1}$, but that requires multiple steps of identity manipulation and is prone to errors. Calculus is the 'universal key'—it works for $r^2, r^3$, and even higher powers. Mastering this method makes you faster and more accurate.

Q: Why do we set $x=1$ at the end?

Look at the original series: $1^2 \binom{20}{1} + 2^2 \binom{20}{2} + \dots$. There are no $x$ terms in the final sum. Setting $x=1$ is the standard way to 'collapse' the variable part of the binomial expansion, leaving you with just the sum of the coefficients. It is the final step that bridges the gap between the algebraic identity and the numerical answer.

Question

If $^{20}C_1 + (2^2) ^{20}C_2 + (3^2) ^{20}C_3 + \dots + (20^2) ^{20}C_{20} = A(2^\beta)$, then the ordered pair $(A, \beta)$ is equal to:

Options:

(A) (420, 18)

(B) (380, 19)

(C) (380, 18)

(D) (420, 19)

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a massive, intimidating wall of numbers. You see $1^2 \binom{20}{1} + 2^2 \binom{20}{2} + \dots + 20^2 \binom{20}{20}$.

At first glance, it looks like a chaotic mess of squares and combinations. But in the world of JEE Advanced, chaos is just order waiting to be discovered.

Today, we are going to peel back the layers of this series using the most elegant tool in our mathematical toolkit: **Calculus**.

### The Foundation: The Binomial Identity

Every great journey starts with a simple truth. We know the binomial expansion:

$$ (1+x)^n = \sum_{r=0}^{n} \binom{n}{r} x^r $$

This is our bedrock. It is a beautiful, symmetric identity.

But notice the difference between this identity and our problem. Our problem has an $r^2$ multiplier, while the identity has a plain $x^r$.

How do we transform $x^r$ into $r^2 x^r$? The answer lies in the power rule of differentiation. When we differentiate $x^r$, we get $r x^{r-1}$. That single $r$ is exactly what we need.

### The Calculus Operator: The First Step

Let us take the derivative of both sides with respect to $x$:

$$ \frac{d}{dx} (1+x)^n = \frac{d}{dx} \sum_{r=0}^{n} \binom{n}{r} x^r $$

On the left, we get $n(1+x)^{n-1}$. On the right, the derivative of $x^r$ is $r x^{r-1}$. So, we have:

$$ n(1+x)^{n-1} = \sum_{r=1}^{n} r \binom{n}{r} x^{r-1} $$

We have successfully brought down one $r$! But wait—we need $r^2$. If we differentiate again right now, we will get $r(r-1)x^{r-2}$, which introduces an unwanted $r$ term. We need to 'reset' the power of $x$ back to $x^r$ before the second differentiation.

### The Reset: Multiplying by $x$

This is the moment where most students stumble, but you will not. To get back to $x^r$, we simply multiply both sides of our equation by $x$:

$$ x \cdot n(1+x)^{n-1} = x \cdot \sum_{r=1}^{n} r \binom{n}{r} x^{r-1} $$

This simplifies beautifully to:

$$ nx(1+x)^{n-1} = \sum_{r=1}^{n} r \binom{n}{r} x^r $$

Now, look at the right side. The power of $x$ is $r$. We are perfectly set up for the second differentiation.

### The Second Derivative: Generating $r^2$

Now, we differentiate with respect to $x$ one more time. On the left side, we must use the product rule because we have $x$ multiplied by $(1+x)^{n-1}$.

The derivative of $nx(1+x)^{n-1}$ is:

$$ n(1+x)^{n-1} + nx(n-1)(1+x)^{n-2} $$

On the right side, the derivative of $x^r$ is $r x^{r-1}$, which gives us:

$$ \sum_{r=1}^{n} r^2 \binom{n}{r} x^{r-1} $$

We have done it! We have isolated the $r^2$ term.

### The Final Evaluation: Setting $x=1$

The variable $x$ was our scaffolding; now that the building is complete, we can take it down. By setting $x=1$, the right side becomes our original series sum, $S$.

On the left side, we substitute $x=1$:

$$ n(1+1)^{n-1} + n(1)(n-1)(1+1)^{n-2} = S $$

$$ n(2)^{n-1} + n(n-1)(2)^{n-2} = S $$

To simplify this, factor out $n \cdot 2^{n-2}$:

$$ S = n \cdot 2^{n-2} [2 + (n-1)] = n(n+1)2^{n-2} $$

This is the general formula for $\sum r^2 \binom{n}{r}$. It is a powerful result that you should keep in your mental library.

### Conclusion: The Final Calculation

For our specific problem, $n=20$. Plugging this into our formula:

$$ S = 20(20+1)2^{20-2} = 20(21)2^{18} = 420 \cdot 2^{18} $$

Comparing this to $A(2^\beta)$, we find **$A=420$** and **$\beta=18$**.

You have navigated the calculus, mastered the operator method, and arrived at the solution. This is the essence of JEE Advanced: not just calculating, but understanding the flow of the math. Keep this logic close, and no series will ever intimidate you again.

Animated Solution for Mathematics - Binomial Theorem: If $^{20} C_{1} + (2^{2})^{20} C_{2} + (3^{2})^{20} C_{3} + \dots + (2 0^{2})^{20} C_{20} = A (2^{β})$ , then the ordered pair $(A, β)$ is equal to:

Select Answer:

Visualized Solution

Identify the Series

The Binomial Expansion

First Differentiation

Multiply by $x$

Second Differentiation

Substitute $x = 1$

Simplify the Sum Formula

Substitute $n = 20$

Calculate Final Values

Final Answer

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The Foundation

The Calculus Operator

The Reset

The Second Derivative

The Final Evaluation

Conclusion

Similar Questions

If $1^{2} \cdot (^{15} C_{1}) + 2^{2} \cdot (^{15} C_{2}) + 3^{2} \cdot (^{15} C_{3}) + \dots + 1 5^{2} \cdot (^{15} C_{15}) = 2^{m} \cdot 3^{n} \cdot 5^{k}$ , where $m, n, k \in N$ , then $m + n + k$ is equal to :

Let $(1 + x + 2 x^{2})^{20} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{40} x^{40}$ , then $a_{1} + a_{3} + a_{5} + \dots + a_{37}$ is equal to

The value of $(^{21} C_{1} -^{10} C_{1}) + (^{21} C_{2} -^{10} C_{2}) + (^{21} C_{3} -^{10} C_{3}) + (^{21} C_{4} -^{10} C_{4}) + \dots + (^{21} C_{10} -^{10} C_{10})$ is:

The number of ordered pairs $(r, k)$ for which $6 \cdot^{35} C_{r} = (k^{2} - 3) \cdot^{36} C_{r + 1}$ , where $k$ is an integer, is :

The number of ordered pairs $(r, k)$ for which $6 .^{35} C_{r} = (k^{2} - 3) .^{36} C_{r + 1}$ , where $k$ is an integer, is:

Let $(k n)$ denotes $^{n} C_{k}$ and $[n k] = {(k n), 0, if 0 \leq k \leq n otherwise$ . If $A_{k} = \sum_{i = 0}^{9} (i 9) [12 12 - k + i] + \sum_{i = 0}^{8} (i 8) [13 13 - k + i]$ and $A_{4} - A_{3} = 190 p$ , then $p$ is equal to :

Let $S_{n} = 1 + q + q^{2} + \dots + q^{n}$ and $T_{n} = 1 + (\frac{q + 1}{2}) + (\frac{q + 1}{2})^{2} + \dots + (\frac{q + 1}{2})^{n}$ where $q$ is a real number and $q \neq = 1$ . If $^{101} C_{1} +^{101} C_{2} \cdot S_{1} + \dots +^{101} C_{101} \cdot S_{100} = α T_{100}$ then $α$ is equal to:

$\sum_{k = 0}^{20} (^{20} C_{k})^{2}$ is equal to :

If $\sum_{r = 1}^{9} (\frac{r + 3}{2 ^{r}}) \cdot^{9} C_{r} = α (\frac{3}{2})^{9} - β, α, β \in N$ , then $(α + β)^{2}$ is equal to

Animated Solution for Mathematics - Binomial Theorem: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 20C1​+(22)20C2​+(32)20C3​+⋯+(202)20C20​=A(2β), then the ordered pair (A,β) is equal to: