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MHTCET 25 Apr 2025 Shift 1 Paper

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Question : 134 of 150

Marks: +1, -0

If

{ }^{n} C_{0}+\frac{1}{2}{ }^{n} C_{1}+\frac{1}{3}{ }^{n} C_{2}+\ldots \frac{1}{n}{ }^{n} C_{n-1}+\frac{1}{n+1}{ }^{n} C_{n}=\frac{1023}{10}

\mathrm{n}=

7
8
9
10

Solution:

Concept:

Use the binomial coefficient identity to convert the given sum into a simpler form.

Explanation:

The given series is

\sum_{r=0}^{n} \frac{1}{r+1} { }^{n} C_{r}

.

Using

{ }^{n+1} C_{r+1} = \frac{n+1}{r+1} { }^{n} C_{r}

, we get

\frac{1}{r+1} { }^{n} C_{r} = \frac{1}{n+1} { }^{n+1} C_{r+1}

.

So the sum becomes

\frac{1}{n+1} \sum_{r=0}^{n} { }^{n+1} C_{r+1}

.

Now

\sum_{r=0}^{n} { }^{n+1} C_{r+1} = 2^{n+1} - 1

because the full binomial sum is

2^{n+1}

and

{ }^{n+1} C_{0} = 1

is missing.

Therefore,

\frac{2^{n+1} - 1}{n+1} = \frac{1023}{10}

.

Testing values,

2^{10} - 1 = 1023

and

n+1 = 10

, so

n = 9

.

Answer:

n = 9

Correct option: C

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