NIMCET 2010 — Mathematics PYQ

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Question

NIMCET 2010 — Mathematics PYQ

NIMCET | Mathematics | 2010

If $(1 + x)^n = a_0 + a_1 x + a_2 x^2 + \dots + a_n x^n$ , then $\left(1 + \frac{a_1}{a_0}\right) \left(1 + \frac{a_2}{a_1}\right) \left(1 + \frac{a_3}{a_2}\right) \dots \left(1 + \frac{a_n}{a_{n-1}}\right)$ is equal to:

Choose the correct answer:

Explanation

Solution

1. Identify the Coefficients

From the binomial expansion of $(1 + x)^n$ , the coefficients $a_r$ are given by:

a_r = ^nC_r = \frac{n!}{r!(n-r)!}

2. Simplify the General Term

Let's find the expression for a general term in the product, $\left(1 + \frac{a_r}{a_{r-1}}\right)$ :

\frac{a_r}{a_{r-1}} = \frac{^nC_r}{^nC_{r-1}}

Using the property $\frac{^nC_r}{^nC_{r-1}} = \frac{n-r+1}{r}$ :

1 + \frac{a_r}{a_{r-1}} = 1 + \frac{n-r+1}{r}

1 + \frac{a_r}{a_{r-1}} = \frac{r + n - r + 1}{r}

1 + \frac{a_r}{a_{r-1}} = \frac{n+1}{r}

3. Evaluate the Product

The given expression is the product of these terms from $r=1$ to $r=n$ :

\text{Product} = \left(\frac{n+1}{1}\right) \left(\frac{n+1}{2}\right) \left(\frac{n+1}{3}\right) \dots \left(\frac{n+1}{n}\right)

4. Final Calculation

The numerator consists of $(n+1)$ multiplied $n$ times, and the denominator is the product of integers from $1$ to $n$ :

\text{Product} = \frac{(n+1)^n}{1 \cdot 2 \cdot 3 \dots n}

\text{Product} = \frac{(n+1)^n}{n!}

Correct Option:

The correct option is (b) $\frac{(n+1)^n}{n!}$ .