NIMCET 2007 — Mathematics PYQ

Step 1: Write down the General Binomial Expansion

The standard binomial expansion for $(1+x{)}^n$ is:

$$(1+x{)}^n=C_0+C_{1}x+C_2{x}^2+C_3{x}^3+\cdots +C_n{x}^n$$

Given that the last term contains $C_{10}$, we choose $n=10$:

$$(1+x{)}^{10}=C_0+C_{1}x+C_2{x}^2+C_3{x}^3+\cdots +C_{10}{x}^{10}$$

Step 2: Integrate Both Sides

To generate the denominators ($2,3,4,\dots$) present in our target expression, we integrate both sides with respect to $x$ from $0$ to $2$:

$${\int }_0^2(1+x{)}^{10}dx={\int }_0^2\left(C_0+C_{1}x+C_2{x}^2+\cdots +C_{10}{x}^{10}\right)dx$$

Step 3: Evaluate the Right-Hand Side (RHS)

Integrate term by term:

$$\text{RHS}={\left[C_{0}x+C_1\frac{x^2}{2}+C_2\frac{x^3}{3}+\cdots +C_{10}\frac{x^{11}}{11}\right]}_0^2$$

Substitute the upper limit $x=2$:

$${\text{RHS}}_{\text{upper}}=C_0(2)+C_1\frac{2^2}{2}+C_2\frac{2^3}{3}+\cdots +C_{10}\frac{2^{11}}{11}$$

$${\text{RHS}}_{\text{upper}}=2C_0+\frac{2^2}{2}{C}_1+\frac{2^3}{3}{C}_2+\cdots +\frac{2^{11}}{11}{C}_{10}$$

Substitute the lower limit $x=0$, which yields $0$. Thus, the RHS perfectly matches our target series.

Step 4: Evaluate the Left-Hand Side (LHS)

Now compute the integral value of the left side:

$$\text{LHS}={\int }_0^2(1+x{)}^{10}dx$$

$$\text{LHS}={\left[\frac{(1+x{)}^{11}}{11}\right]}_0^2$$

Apply the limits:

$$\text{LHS}=\frac{(1+2{)}^{11}}{11}-\frac{(1+0{)}^{11}}{11}$$

$$\text{LHS}=\frac{3^{11}}{11}-\frac{1^{11}}{11}$$

$$\text{LHS}=\frac{3^{11}-1}{11}$$

Conclusion

The sum of the series is equal to $\frac{3^{11}-1}{11}$.

Correct Option: (a) $\frac{3^{11}-1}{11}$