SATHEE: Binomial Theorem 1 Question 4

Binomial Theorem 1 Question 4

5.

If some three consecutive coefficients in the binomial expansion of $(x + 1)^{n}$ in powers of $x$ are in the ratio $2 : 15$ : 70 , then the average of these three coefficients is

(2019 Main, 9 April II)

(a) 964

(b) 227

(d) 625

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Answer:

Correct Answer: 5. (c)

Solution:

Key Idea: Use general term of Binomial expansion $(x + a)^{n}$ i.e.

$T_{r + 1} =^{n} C_{r 1} x^{n - r} a^{r}$

Given binomial is $(x + 1)^{n}$ , whose general term, is $T_{r + 1} =^{n} C_{r} x^{r}$

According to the question, we have

$^{n} C_{r - 1} :^{n} C_{r} :^{n} C_{r + 1} = 2 : 15 : 70$

Now, $\frac{^{n} C_{r - 1}}{^{n} C_{r}} = \frac{2}{15}$

$\Rightarrow \frac{\frac{n!}{(r - 1)! (n - r + 1)!}}{\frac{n!}{r! (n - r)!}} = \frac{2}{15}$ $\Rightarrow \frac{r}{n - r + 1} = \frac{2}{15}$

$\Rightarrow 15 r = 2 n - 2 r + 2$

$\Rightarrow 2 n - 17 r + 2 = 0$ …….(i)

Similarly, $\frac{^{n} C_{r}}{^{n} C_{r + 1}} = \frac{15}{70}$

$\Rightarrow \frac{\frac{n!}{r! (n - r)!}}{\frac{n!}{(r - 1)! (n - r - 1)!}} = \frac{3}{14}$

$\Rightarrow \frac{r + 1}{n - r} = \frac{3}{14} \Rightarrow 14 r + 14 = 3 n - 3 r$

$\Rightarrow 3 n - 17 r - 14 = 0$ …….(ii)

On solving Eqs. (i) and (ii), we get

$n - 16 = 0 \Rightarrow n = 16$ and $r = 2$

Now, the average $= \frac{^{16} C_{1} +^{16} C_{2} +^{16} C_{3}}{3}$

$= \frac{16 + 120 + 560}{3} = \frac{696}{3} = 232$

Binomial Theorem 1 Question 4

5.

Answer:

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Binomial Theorem 1 Question 4

5.

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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