Question

If the r^th term in the expansion of \(\left( \dfrac{x}{3} - \dfrac{2}{x^2} \right)^{10}\)contains x⁴, then r is equal to
1. 3
2. 4
3. 2
4. None of these

Answer 1

Correct Answer - Option 1 : 3

Concept: 

The general term in Binomial Expansion :

The binomial expansion of (x + y)ⁿ,

(x+ y)ⁿ = ⁿC₀( xⁿ) + ⁿC₁ (x^{n - 1}) y + ⁿC(x^{n- 2}). y² +....... + nC_nyⁿ 

General term of binomial expansion = T_r+1 = ⁿ C _r ( x) ^{n - r} (a) ^r in the expansion of (x + a)ⁿ

Calculation: 

We need to find which term contains the 4th power of x in the binomial expansion of \(\left( \dfrac{x}{3} - \dfrac{2}{x^2} \right)^{10}\)

Method 1:

T_r in the expansion of \(\left( \dfrac{x}{3} - \dfrac{2}{x^2} \right)^{10}\)

⇒ ¹⁰C_{r - 1} (x / 3)^{10 - ( r -1 )} (- 2 / x²)^{r - 1}  ---(1)

⇒ 10C_{r - 1} (x)^{13 - 3r} (3)^{-11 + r} (- 1)^r(2)^{r - 1}

For x⁴ ,

⇒ 13 - 3r = 4

⇒ 3r = 9

⇒ r = 3 

Method 2:

Let us do the conventional binomial expansion and verify the answer,

\(\left( \dfrac{x}{3} - \dfrac{2}{x^2} \right)^{10} = 10 \left(\frac{x}{3}\right)^{10}+10\left(\frac{x}{3}\right)^{10-1} \left(\frac{-2}{x^2}\right)^{1}+45\left(\frac{x}{3}\right)^{10-2} \left(\frac{-2}{x^2}\right)^{2}+....\)

Clearly, we can see that,

1st term → contains x¹⁰

2nd term → contains x(9-2) = x⁷

3rd term → contains x^(8-4) = x⁴

The 3rd term in the binomial expansion contains the 4^th power of x.

Do not get confused with the (r - 1) in the formula for the r^th term (equation 1). 

It does not represent the (r - 1)^th term.