Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# If in the Expansion of (1 + X)N, the Coefficients of Three Consecutive Terms Are 56, 70 and 56, Then Find N and the Position of the Terms of These Coefficients. - Mathematics

If in the expansion of (1 + x)n, the coefficients of three consecutive terms are 56, 70 and 56, then find n and the position of the terms of these coefficients.

#### Solution

$\text{ Suppose } r^{th} , (r + 1) {}^{th} \text{ and } (r + 2 )^{th} \text{ terms are the three consecutive terms .}$

$\text{ Their respective coefficients are } ^{n}{}{C}_{r - 1} , ^{n}{}{C}_r \text{ and } ^{n}{}{C}_{r + 1} .$

$\text{ We have: }$

$^{n}{}{C}_{r - 1} =^{n}{}{C}_{r + 1} = 56$

$\Rightarrow r - 1 + r + 1 = n [\text{ If } ^{n}{}{C}_r = ^{n}{}{C}_s \Rightarrow r = s \text{ or } r + s = n]$

$\Rightarrow 2r = n$

$\Rightarrow r = \frac{n}{2}$

$\text{ Now } ,$

$^{n}{}{C}_\frac{n}{2} = 70 \text{ and }^{n}{}{C}_\left( \frac{n}{2} - 1 \right) = 56$

$\Rightarrow \frac{^{n}{}{C}_\left( \frac{n}{2} - 1 \right)}{^{n}{}{C}_\frac{n}{2}} = \frac{56}{70}$

$\Rightarrow \frac{\frac{n}{2}}{\left( \frac{n}{2} + 1 \right)} = \frac{8}{10}$

$\Rightarrow 5n = 4n + 8$

$\Rightarrow n = 8$

$So, r = \frac{n}{2} = 4$

$\text{ Thus, the required terms are 4th, 5th and 6th .}$

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 18 Binomial Theorem
Exercise 18.2 | Q 28 | Page 40

Share