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Q.
In the expansion of $(1 + x)^{18}$ , if the coefficients of $(2r + 4)^{th}$ and $(r - 2)^{th}$ terms are equal, then the value of r is :
Binomial Theorem
Solution:
Since the coefficient of $(r +1)^{th}$ term in the expansion of $\left(1 + x\right)^{n} = ^{n}C_{r}$
$\therefore $ In the expansion of $\left(1+ x\right)^{18}$
coefficient of $\left(2r + 4\right)^{th}$ term $= ^{18}C_{2r + 3}$,
Similarly, coefficient of $\left(r-2\right)^{th}$ term in the expansion of $\left(1+ x\right)^{18} = ^{18}C_{r-3}$
If $^{n}C_{r} = ^{n}C_{s}$ then $r + s = n$
So, $^{18}C_{2r+3} = ^{18}C_{r-3}$ gives
$2r + 3 + r - 3 = 18$
$\Rightarrow 3r = 18 \Rightarrow r = 6.$