Find the value of $r$ if the coefficients of $(2 r + 4)$ th and $(r - 2) t h$ terms in the expansion of $(1 + x)^{18}$ are equal.

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Solution

$(2 r + 4) t h$ term $= {^{18} C}_{2 r + 3} x^{2 r + 3}$
$\Rightarrow (r - 2) t h$ term $= {^{18} C}_{r - 3} x^{r - 3}$
So, as coefficient are equal
$\Rightarrow {^{18} C}_{2 r + 3} = {^{18} C}_{r - 3}$
$\Rightarrow 2 r + 3 = r - 3$ $15 - 2 r = r - 3$
$r = - 6$ $18 = 3 r$
$r = - 6$ $r = 6$
Not possible
So, $r = 6.$
Hence, the answer is $6.$

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