SATHEE: Binomial Theorem 1 Question 34

Binomial Theorem 1 Question 34

36.

Let $m$ be the smallest positive integer such that the coefficient of $x^{2}$ in the expansion of $(1 + x)^{2} + (1 + x)^{3} + \dots + (1 + x)^{49} + (1 + m x)^{50}$ is $(3 n + 1)$ $^{51} C_{3}$ for some positive integer $n$ . Then, the value of $n$ is

(2016 Adv.)

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Answer:

Correct Answer: 36. $(5)$

Solution:

Coefficient of $x^{2}$ in the expansion of

$(1 + x)^{2} + (1 + x)^{3} + \dots + (1 + x)^{49} + (1 + m x)^{50}$

$\Rightarrow^{2} C_{2} +^{3} C_{2} +^{4} C_{2} + \dots +^{49} C_{2} +^{50} C_{2} \cdot m^{2}$

$\begin{array}{r} \Rightarrow^{2} C_{2} +^{3} C_{2} +^{4} C_{2} + \dots +^{49} C_{2} +^{50} C_{2} \cdot m^{2} \\ = (3 n + 1) \cdot^{51} C_{3} \end{array}$

$\Rightarrow^{50} C_{3} +^{50} C_{2} m^{2} = (3 n + 1) \cdot^{51} C_{3}$

$[∵^{r} C_{r} +^{r + 1} C_{r} + \dots +^{n} C_{r} =^{n + 1} C_{r + 1}]$

$\Rightarrow \frac{50 \times 49 \times 48}{3 \times 2 \times 1} + \frac{50 \times 49}{2} \times m^{2} = (3 n + 1) \frac{51 \times 50 \times 49}{3 \times 2 \times 1}$

$\Rightarrow m^{2} = 51 n + 1$

$∴$ Minimum value of $m^{2}$ for which $(51 n + 1)$ is integer (perfect square) for $n = 5$ .

$\begin{array}{ll} ∴ & m^{2} = 51 \times 5 + 1 \Rightarrow m^{2} = 256 \\ ∴ & m = 16 and n = 5 \end{array}$

Hence, the value of $n$ is $5$ .

Binomial Theorem 1 Question 34

36.

Answer:

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Binomial Theorem 1 Question 34

36.

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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