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

We have,

$(1 + x)^{n}$

Coefficients of pth term $=^{n} C_{p - 1}$

and, Coefficients of qth term $=^{n} C_{q - 1}$

It is given that, these coefficients are equal.

$∴^{n} C_{p - 1} =^{n} C_{q - 1}$

$\Rightarrow p - 1 = q - 1$ or, p-1 +q-1 =n

$[∵^{n} C_{r} =^{n} C_{s} \Rightarrow r = s$ or r+s =n]

$\Rightarrow p - q = 0$ or, p+q =n+2

$∴ p + q = n + 2$

Hence proved.

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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.

We have, (1+x)n Coefficients of pth term =nCp−1 and, Coefficients of qth term =nCq−1 It is given that, these coefficients are equal. ∴nCp−1=nCq−1 ⇒p−1=q−1 or, p-1 +q-1 =n [∵nCr=nCs⇒r=s or r+s =n] ⇒p−q=0 or, p+q =n+2 ∴p+q=n+2 Hence proved.

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