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Question
If the coefficients of three consecutive terms in the expansion of (1 + x)n be 76, 95 and 76, find n.
Solution
\[\text{ Suppose r, } \left( r + 1 \right) \text{ and } \left( r + 2 \right) \text{ are three consecutive terms in the given expansion } . \]
\[\text { The coefficients of these terms are } ^{n}{}{C}_{r - 1} , ^{n}{}{C}_r \text{ and } ^ {n}{}{C}_{r + 1} . \]
\[\text{ According to the question, } \]
\[ ^{n}{}{C}_{r - 1} = 76\]
\[^{n}{}{C}_r = 95\]
\[ ^{n}{}{C}_{r + 1} = 76\]
\[ \Rightarrow ^{n}{}{C}_{r - 1} = ^{n}{}{C}_{r + 1} \]
\[ \Rightarrow r - 1 + r + 1 = n [\text{ If } ^{n}{}{C}_r = ^{n}{}{C}_s \Rightarrow r = s \text{ or } r + s = n]\]
\[ \Rightarrow r = \frac{n}{2}\]
\[ \therefore \frac{^{n}{}{C}_r}{^{n}{}{C}_{r - 1}} = \frac{95}{76}\]
\[ \Rightarrow \frac{n - r + 1}{r} = \frac{95}{76}\]
\[ \Rightarrow \frac{\frac{n}{2} + 1}{\frac{n}{2}} = \frac{95}{76}\]
\[ \Rightarrow 38n + 76 = \frac{95n}{2} \]
\[ \Rightarrow \frac{19n}{2} = 76\]
\[ \Rightarrow n = 8\]
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