यदि $(2+a)^{50}$ के द्विपद प्रसार का सत्रहवाँ और अट्ठारहवाँ पद समान हो तो $a$ का मान ज्ञात कीजिए।

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., $a = \frac{{{\,^{50}}{C_{16}} \times 2}}{{{\,^{50}}{C_{17}}}} = \frac{{50!}}{{16!34!}} \times \frac{{17! \cdot 33!}}{{50!}} \times 2 = 1$