SATHEE: Binomial Theorem 1 Question 3

Binomial Theorem 1 Question 3

4.

The smallest natural number $n$ , such that the coefficient of $x$ in the expansion of $(x^{2} + \frac{1}{x^{3}})^{n}$ is $^{n} C_{23}$ , is

(2019 Main, 10 April II)

(a) 35

(b) 23

(d) 38

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Answer:

Correct Answer: 4. (d)

Solution:

Given binomial is $(x^{2} + \frac{1}{x^{3}})^{n}$ , its $(r + 1)^{th}$ term, is

$\begin{aligned} T_{r + 1} & =^{n} C_{r} {(x^{2})}^{n - r} (\frac{1}{x^{3}})^{r} =^{n} C_{r} x^{2 n - 2 r} \frac{1}{x^{3 r}} \\ =^{n} C_{r} x^{2 n - 2 r - 3 r} =^{n} C_{r} x^{2 n - 5 r} \end{aligned}$

For the coefficient of $x$ ,

$2 n - 5 r = 1 \Rightarrow 2 n = 5 r + 1 \dots (i)$

As coefficient of $x$ is given as $^{n} C_{23}$ , then either $r = 23$ or $n - r = 23$ .

If $r = 23$ , then from Eq. (i), we get

$2 n = 5 (23) + 1$

$\Rightarrow 2 n = 115 + 1 \Rightarrow 2 n = 116 \Rightarrow n = 58$ .

If $n - r = 23$ , then from Eq. (i) on replacing the value of $r$ , we get $2 n = 5 (n - 23) + 1$

$\Rightarrow 2 n = 5 n - 115 + 1 \Rightarrow 3 n = 114 \Rightarrow n = 38$

So, the required smallest natural number $n = 38$ .

Binomial Theorem 1 Question 3

4.

Answer:

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Binomial Theorem 1 Question 3

4.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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