NIMCET 2007 — Mathematics PYQ

There are two easy methods to solve this problem. Let's look at both.

Method 1: Using Taylor/Maclaurin Series Expansion

According to Taylor's series expansion for a polynomial function $f(x)$ centered at $a=1$:

$$f(x)=f(1)+\frac{f^{\prime }(1)}{1!}(x-1)+\frac{f^{\prime \prime }(1)}{2!}(x-1{)}^2+\cdots +\frac{f^n(1)}{n!}(x-1{)}^n$$

Now, if we substitute $x=2$ into this series formula, we get:

$$f(2)=f(1)+\frac{f^{\prime }(1)}{1!}(2-1)+\frac{f^{\prime \prime }(1)}{2!}(2-1{)}^2+\cdots +\frac{f^n(1)}{n!}(2-1{)}^n$$

Since $(2-1)=1$, the expression simplifies perfectly to our target equation:

$$f(2)=f(1)+\frac{f^{\prime }(1)}{1!}+\frac{f^{\prime \prime }(1)}{2!}+\cdots +\frac{f^n(1)}{n!}$$

Given the original function is $f(x)=x^n$, we simply plug in $x=2$:

$$f(2)=2^n$$

Therefore, the value of the given expression is $2^n$.

Method 2: Using Derivatives and Binomial Coefficients

Let's find the derivatives of $f(x)=x^n$ at $x=1$:

• $f(1)=1^n=1$ (which is $\left(\genfrac{}{}{0ex}{}{n}{0}\right)$)

• $f^{\prime }(x)=n\cdot x^{n-1}⟹f^{\prime }(1)=n$

• $f^{\prime \prime }(x)=n(n-1)x^{n-2}⟹f^{\prime \prime }(1)=n(n-1)$

• $f^r(1)=n(n-1)\dots (n-r+1)$

Now, let's substitute these derivative values back into our given series:

$$\text{Series}=1+\frac{n}{1!}+\frac{n(n-1)}{2!}+\cdots +\frac{n(n-1)\dots 1}{n!}$$

Notice that these terms are exactly the standard binomial coefficients $\left(\genfrac{}{}{0ex}{}{n}{r}\right)$:

$$\text{Series}=\left(\genfrac{}{}{0ex}{}{n}{0}\right)+\left(\genfrac{}{}{0ex}{}{n}{1}\right)+\left(\genfrac{}{}{0ex}{}{n}{2}\right)+\cdots +\left(\genfrac{}{}{0ex}{}{n}{n}\right)$$

From the properties of binomial coefficients, we know that the sum of all binomial coefficients of order $n$ equals $2^n$:

$$\sum _{r=0}^n\left(\genfrac{}{}{0ex}{}{n}{r}\right)=2^n$$

Conclusion

Both analytical approaches successfully yield the same result.

Final Answer: $2^n$ (Option B)