JEE Main 2022 — Binomial Theorem Question with Solution

Given Binomial Expansion

$$={\left(\frac{\sqrt{x}}{5^{\frac{1}{4}}}+\frac{\sqrt{x}}{5^{\frac{1}{3}}}\right)}^{60}$$

$$\therefore$$ General term

$$T_{r+1}={}^{60}{C}_r.{\left(\frac{x^{1/2}}{5^{1/4}}\right)}^{60-r}.{\left(\frac{5^{1/2}}{x^{1/3}}\right)}^r$$

$$={}^{60}{C}_r.5^{\left(\frac{r}{4}-15+\frac{r}{2}\right)}.x^{\left(30-\frac{r}{2}-\frac{r}{3}\right)}$$

$$={}^{60}{C}_r.5^{\left(\frac{3r-60}{4}\right)}.x^{\left(\frac{180-5r}{6}\right)}$$

For x¹⁰ term,

$$\frac{180-5r}{6}=10$$

$$\Rightarrow 5r=120$$

$$\Rightarrow r=24$$

$$\therefore$$ Coefficient of $$x^{10}={}^{60}{C}_{24}.5^{\left(\frac{3\times 24-60}{4}\right)}$$

$$={}^{60}{C}_{24}.5^3$$

$$=\frac{60!}{24!36!}.5^3$$

It is given that,

$$\frac{60!}{24!36!}.5^3=5^k.l$$ ...... (1)

Also given that, l is coprime to 5 means l can't be multiple of 5. So we have to find all the factors of 5 in 60!, 24! and 36!

[Note : Formula for exponent or degree of prime number in n!.

Exponent of p in $$n!=\lceil \frac{n}{p}\rceil +\lceil \frac{n}{p^2}\rceil +\lceil \frac{n}{p^3}\rceil +$$ ..... until 0 comes

here p is a prime number. ]

$$\therefore$$ Exponent of 5 in 60!

$$=\lceil \frac{60}{5}\rceil +\lceil \frac{60}{5^2}\rceil +\lceil \frac{60}{5^3}\rceil +$$ .....

$$=12+2+0+$$ .....

$$=14$$

Exponent of 5 in 24!

$$=\lceil \frac{24}{5}\rceil +\lceil \frac{24}{5^2}\rceil +\lceil \frac{24}{5^3}\rceil +$$ ......

$$=4+0+0$$ ......

$$=4$$

Exponent of 5 in 36!

$$=\lceil \frac{36}{5}\rceil +\lceil \frac{36}{5^2}\rceil +\lceil \frac{36}{5^3}\rceil +$$ .......

$$=7+1+0$$ ......

$$=8$$

$$\therefore$$ From equation (1), exponent of 5 overall

$$\frac{5^{14}}{5^4.5^8}.5^3=5^k$$

$$\Rightarrow 5^5=5^k$$

$$\Rightarrow k=5$$