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If in the Expansion of (1 + X)N, the Coefficients of Pth and Qth Terms Are Equal, Prove that P + Q = N + 2, Where P ≠ Q

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Question

If in the expansion of (1 + x)n, the coefficients of pth and qth terms are equal, prove that p + q = n + 2, where  \[p \neq q\]

 

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Solution

\[\text{ Coefficients of the pth and qth terms are }  ^{n}{}{C}_{p - 1} \text{ and }^{n}{}{C}_{q - 1} \text{ respectively } . \]
\[\text{ Thus, we have: } \]
\[ ^{n}{}{C}_{p - 1} =^{n}{}{C}_{q - 1} \]
\[ \Rightarrow p - 1 = q - 1 \text{ or,}  p - 1 + q - 1 = n [ \because ^{n}{}{C}_r =^n C_s \Rightarrow r = s \text{ or, }  r + s = n]\]
\[ \Rightarrow p = q\text{ or } , p + q = n + 2\] 

If \[p \neq q\], then \[p + q = n + 2\]

 

 

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Chapter 18: Binomial Theorem - Exercise 18.2 [Page 39]

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R.D. Sharma Mathematics [English] Class 11
Chapter 18 Binomial Theorem
Exercise 18.2 | Q 24 | Page 39

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