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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

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Question

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.

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Solution

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

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

\[\text{ We have: }  \]

\[ ^{n}{}{C}_{r - 1} =^{n}{}{C}_{r + 1} = 56\]

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

\[ \Rightarrow 2r = n\]

\[ \Rightarrow r = \frac{n}{2}\]

\[\text{ Now } , \]

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

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

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

\[ \Rightarrow 5n = 4n + 8\]

\[ \Rightarrow n = 8\]

\[So, r = \frac{n}{2} = 4\]

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

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Chapter 18: Binomial Theorem - Exercise 18.2 [Page 40]

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

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