Question
If $$^n{C_{r - 1}} + {\,^{n + 1}}{C_{r - 1}} + {\,^{n + 2}}{C_{r - 1}} + ..... + {\,^{2n}}{C_{r - 1}} = {\,^{2n + 1}}{C_{{r^2} - 132}} - {\,^n}{C_r},$$ then the value of $$r$$ and the minimum value of $$n$$ are
A.
10
B.
11
C.
12
D.
13
Answer :
12
Solution :
Given, $$^n{C_r} + {\,^n}{C_{r - 1}} + {\,^{n + 1}}{C_{r - 1}} + {\,^{n + 2}}{C_{r - 1}} + ..... + {\,^{2n}}{C_{r - 1}} = {\,^{2n + 1}}{C_{{r^2} - 132}}$$
$$\eqalign{
& \Rightarrow {\,^{n + 1}}{C_r} + {\,^{n + 1}}{C_{r - 1}} + ..... + {\,^{2n}}{C_{r - 1}} = {\,^{2n + 1}}{C_{{r^2} - 132}} \cr
& .................................................. \cr
& \Rightarrow {\,^{2n}}{C_r} + {\,^{2n}}{C_{r - 1}} = {\,^{2n + 1}}{C_{{r^2} - 132}} \cr
& \Rightarrow {\,^{2n + 1}}{C_r} = {\,^{2n + 1}}{C_{{r^2} - 132}} \cr
& \Rightarrow {r^2} - r - 132 = 0 \cr
& \Rightarrow \left( {r - 12} \right)\left( {r + 11} \right) = 0 \cr
& \Rightarrow r = 12 \cr
& \Rightarrow n \geqslant 12 \cr} $$
So, minimum value of $$n = 12.$$