NIMCET 2023 — Mathematics PYQ

Let the given expression be $S$:

$$S=(1+x{)}^{1000}+2x(1+x{)}^{999}+3x^2(1+x{)}^{998}+\cdots +1001x^{1000}$$

This series is an Arithmetico-Geometric Progression (AGP). We can simplify it using the standard shift-and-subtract method.

Step 1: Multiply the entire series by $\frac{x}{1+x}$

$$\frac{x}{1+x}\cdot S=x(1+x{)}^{999}+2x^2(1+x{)}^{998}+\cdots +1000x^{1000}+\frac{1001x^{1001}}{1+x}$$

Step 2: Subtract this new equation from the original equation ($S$)

$$S-\frac{x}{1+x}\cdot S=(1+x{)}^{1000}+\left[2x(1+x{)}^{999}-x(1+x{)}^{999}\right]+\left[3x^2(1+x{)}^{998}-2x^2(1+x{)}^{998}\right]+\cdots -\frac{1001x^{1001}}{1+x}$$

$$S\left(1-\frac{x}{1+x}\right)=(1+x{)}^{1000}+x(1+x{)}^{999}+x^2(1+x{)}^{998}+\cdots +x^{1000}-\frac{1001x^{1001}}{1+x}$$

$$S\left(\frac{1}{1+x}\right)=\left[(1+x{)}^{1000}+x(1+x{)}^{999}+x^2(1+x{)}^{998}+\cdots +x^{1000}\right]-\frac{1001x^{1001}}{1+x}$$

Step 3: Simplify the Geometric Progression (GP) inside the brackets

The terms inside the bracket form a GP where:

• First term ($a$) = $(1+x{)}^{1000}$

• Common ratio ($r$) = $\frac{x}{1+x}$

• Number of terms ($n$) = $1001$

Using the GP sum formula $S_n=\frac{a(1-r^n)}{1-r}$:

$$\text{Sum}=\frac{(1+x{)}^{1000}\left[1-{\left(\frac{x}{1+x}\right)}^{1001}\right]}{1-\frac{x}{1+x}}$$

$$\text{Sum}=\frac{(1+x{)}^{1000}\cdot \frac{(1+x{)}^{1001}-x^{1001}}{(1+x{)}^{1001}}}{\frac{1}{1+x}}$$

$$\text{Sum}=(1+x)\cdot \frac{(1+x{)}^{1001}-x^{1001}}{1+x}=(1+x{)}^{1001}-x^{1001}$$

Step 4: Substitute the GP sum back into the main equation

$$S\left(\frac{1}{1+x}\right)=\left[(1+x{)}^{1001}-x^{1001}\right]-\frac{1001x^{1001}}{1+x}$$

Now, multiply both sides by $(1+x)$ to isolate $S$:

$$S=(1+x{)}^{1002}-x^{1001}(1+x)-1001x^{1001}$$

$$S=(1+x{)}^{1002}-x^{1001}-x^{1002}-1001x^{1001}$$

$$S=(1+x{)}^{1002}-1002x^{1001}-x^{1002}$$

Step 5: Find the coefficient of $x^{50}$

We need the coefficient of $x^{50}$ in the simplified expression:

$$S=(1+x{)}^{1002}-1002x^{1001}-x^{1002}$$

• The terms $1002x^{1001}$ and $x^{1002}$ have powers of $x$ much higher than $50$, so they contribute $0$ to the coefficient of $x^{50}$.

• Therefore, the coefficient of $x^{50}$ comes solely from $(1+x{)}^{1002}$.

Using the general term formula for $(1+x{)}^n$, which is $T_{r+1}=\left(\genfrac{}{}{0ex}{}{n}{r}\right)x^r$:

$$\text{Coefficient of }{x}^{50}=\left(\genfrac{}{}{0ex}{}{1002}{50}\right)={}^{1002}{C}_{50}$$

Correct Answer: C) ${}^{1002}{C}_{50}$