Question

JEE Mains Maths question is in the image

Answer 1

Correct Question:-

If the expansion of $\left(2^{\frac{1}{2}}+5^{\frac{1}{8}}\right)^n$ contains 33 integral terms, find the least value of $n.$

Solution:-

The $(r+1)^{th}$ term of the expansion is,

$\displaystyle\longrightarrow T_{r+1}=\,^n\!C_r\,\left(2^{\frac{1}{2}}\right)^{n-r}\,\left(5^{\frac{1}{8}}\right)^r$

$\displaystyle\longrightarrow T_{r+1}=\,^n\!C_r\cdot2^{\frac{n-r}{2}}\cdot5^{\frac{r}{8}}$

From the expansion we get that $r$ should be exactly divisible by 8 else $5^{\frac{r}{8}}$ will not be an integer, so will be the term.

By the expansion we get last term contains $5^{\frac{n}{8}}.$

First integral term in the expansion is $T_1$ which contains $5^0.$

Next integral term is $T_9$ which contains $5^1.$

Next is $T_{17}$ which contains $5^2.$

Likewise the integral term in the expansion is in the form $T_{8k+1}$ which contains $5^k,$ where $k$ is a whole number.

If the expansion contains 33 integral terms, the expansion should contain terms including $5^0,\ 5^1,\ 5^2,\,\dots\,,5^{32}$ each.

For least $n$ the last term should contain $5^{32}.$

$\Longrightarrow \dfrac{n}{8}=32$

$\longrightarrow\underline{\underline{n=256}}$

Hence (1) is the answer.