Find $a$ if the $17^{\text {th }}$ and $18^{\text {th }}$ terms of the expansion ${(2 + a)^{{\rm{50 }}}}$ are equal.
Find $a$ if the $17^{\text {th }}$ and $18^{\text {th }}$ terms of the expansion ${(2 + a)^{{\rm{50 }}}}$ are equal.
The $(r+1)^{\text {th }}$ term of the expansion $(x+y)^{n}$ is given by ${T_{r + 1}} = \,{\,^n}{C_r}{x^{n - r}}{y^r}$
For the $17^{\text {th }}$ term, we have, $r+1=17,$ i.e., $r=16$
Therefore, ${T_{17}} = {T_{16 + 1}} = {\,^{50}}{C_{16}}{(2)^{50 - 16}}{a^{16}}$
$ = {\,^{50}}{C_{16}}{2^{34}}{a^{16}}$
Simlarly, ${T_{18}} = {\,^{50}}{C_{17}}{2^{33}}{a^{17}}$
Given that $T_{17}=T_{18}$
So ${\,^{50}}{C_{16}}{(2)^{34}}{a^{16}} = {\,^{50}}{C_{17}}{(2)^{33}}{a^{17}}$
Therefore $\frac{{{\,^{50}}{C_{16}} \cdot {2^{34}}}}{{{\,^{50}}{C_{17}} \cdot {2^{33}}}} = \frac{{{a^{17}}}}{{{a^{16}}}}$
i.e., $a = \frac{{{\,^{50}}{C_{16}} \times 2}}{{{\,^{50}}{C_{17}}}} = \frac{{50!}}{{16!34!}} \times \frac{{17! \cdot 33!}}{{50!}} \times 2 = 1$
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