Set power of x = 0 and solve for r. Classic JEE question.

Find the Term Independent of x in (x² + 1/x)⁹

Question

Find the term independent of $x$ in the expansion of $\left(x^2 + \dfrac{1}{x}\right)^9$ .

Solution — Step by Step

The general term in the binomial expansion of $(a + b)^n$ is:

T_{r+1} = \binom{n}{r} \cdot a^{n-r} \cdot b^r

Here $a = x^2$ , $b = \dfrac{1}{x}$ , and $n = 9$ . Substituting:

T_{r+1} = \binom{9}{r} \cdot (x^2)^{9-r} \cdot \left(\frac{1}{x}\right)^r

We need to collect all the $x$ terms in one place before we can do anything useful.

T_{r+1} = \binom{9}{r} \cdot x^{2(9-r)} \cdot x^{-r} = \binom{9}{r} \cdot x^{18 - 2r - r}

T_{r+1} = \binom{9}{r} \cdot x^{18 - 3r}

“Independent of $x$ ” simply means the power of $x$ is zero — that term is a pure constant.

18 - 3r = 0 \implies r = 6

Plug $r = 6$ back into the general term:

T_7 = \binom{9}{6} \cdot x^0 = \binom{9}{6}

\binom{9}{6} = \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84

The term independent of $x$ is $84$ .

Why This Works

Every term in a binomial expansion is a product of two powers — one from each part of the binomial. The $r$ controls how much of each part we use. When we write the general term, we’re really asking: “for this particular split ( $r$ from one part, $n-r$ from the other), what does the power of $x$ come out to?”

Setting the exponent equal to zero is just algebra from there. We’re finding which value of $r$ makes the $x$ disappear completely, leaving only the numerical coefficient $\binom{9}{r}$ .

This is a standard JEE technique — once you can write $T_{r+1}$ cleanly and read off the exponent of $x$ , the rest is just solving a linear equation. The whole skill is in Step 2: collecting exponents correctly.

Alternative Method — Direct Expansion Thinking

Rather than using the formula mechanically, think about what combination gives $x^0$ .

Each term picks from two choices: take $x^2$ (contributing $+2$ to the exponent) or take $\frac{1}{x}$ (contributing $-1$ ). Over 9 picks, if we choose $x^2$ exactly $(9-r)$ times and $\frac{1}{x}$ exactly $r$ times:

2(9 - r) - r = 0 \implies 18 - 3r = 0 \implies r = 6

So we pick $\frac{1}{x}$ six times and $x^2$ three times. The number of ways to do this is $\binom{9}{6} = 84$ .

This “contribution” thinking is faster in MCQs. Once you write down the two exponent values ( $+2$ and $-1$ here), you can mentally set up $2(9-r) - r = 0$ without writing the full general term.

Common Mistake

The most common error is writing the general term as $T_{r+1} = \binom{9}{r}(x^2)^r \cdot \left(\frac{1}{x}\right)^{9-r}$ — putting $r$ on the wrong term. This gives the exponent as $2r - (9-r) = 3r - 9$ , leading to $r = 3$ and the answer $\binom{9}{3} = 84$ . Coincidentally, you still get 84 here because $\binom{9}{3} = \binom{9}{6}$ , but the reasoning is wrong. In questions where the two binomial terms are not “complementary” like this, you will get a completely wrong answer. Always be consistent — if you use $r$ for the second term, use $9-r$ for the first.

Question

Solution — Step by Step

Why This Works

Alternative Method — Direct Expansion Thinking

Common Mistake

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