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7.Binomial Theorem
medium
यदि $(1+x)^{34}$ के प्रसार में $(r-5)^{th}$ और$(2 r-1)^{th}$ पदों के गुणांक समान हों $r$ ज्ञात कीजिए।
A
$14$
B
$14$
C
$14$
D
$14$
Solution
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, $r=14$
Standard 11
Mathematics
