If the number of integral terms in the expansion
of (3^1/2 + 5^1/8)ⁿ is exactly 33, then the least value of n is :

General term of the expression,<br><br>$${T_{r + 1}} = {}^n{C_r}{\left( {{3^{{1 \over 2}}}} \right)^{n - r}}{\left( {{5^{{1 \over 8}}}} \right)^r}$$<br><br>$$ = {}^n{C_r}{\left( 3 \right)^{{{n - r} \over 2}}}{\left( 5 \right)^{{r \over 8}}}$$<br><br>We will get integral term when ${{n - r} \over 2}$ and ${r \over 8}$ are integer<br><br>$\therefore$ <b>(1)</b> n $-$ r is multiple of 2<br><br>$\Rightarrow$ n $-$ r = 0, 2, 4, ......<br><br><b>(2)</b> r is multiple of 8<br><br>$\Rightarrow$ r = 0, 8, 16, .......<br><br>From this two conditions common values are = 0, 8, 16, ....... which will becomes integral terms.<br><br>Given that there are 33 integral terms.<br><br>Here first integral term at 0^th position.<br><br>Second integral term at 8^th position.<br><br>$\therefore$ 33^th integral term will be at = 0 + (33 $-$ 1)8 = 256<br><br>So, there should be at least 256 terms.