If the coefficients of $(r-5)^{th}$ and $(2 r-1)^{th}$ terms in the expansion of $(1+x)^{34}$ are equal, find $r$
If the coefficients of $(r-5)^{th}$ and $(2 r-1)^{th}$ terms in the expansion of $(1+x)^{34}$ are equal, find $r$
The coefficients of $(r-5)^{ th }$ and $(2 r-1)^{th }$ terms of the expansion $(1+x)^{34}$ are $^{34}{C_{r - 6}}$ and $^{34}{C_{2r - 2}},$ respectively. Since they are equal so ${\,^{34}}{C_{r - 6}} = {\,^{34}}{C_{2r - 2}}$
Therefore, either $r-6=2 r-2$ or $r-6=34-(2 r-2)$
[Using the fact that if ${\,^n}{C_r} = {\,^m}{C_p},$ then either $r = p$ or $r = n - p$ ]
So, we get $r=-4$ or $r=14 . r$ being a natural number, $r=-4$ is not possible. So, $r=14$
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