Tᵣ₊₁ = ¹⁰Cᵣ x¹⁰⁻ʳ (1/x)ʳ = ¹⁰Cᵣ x¹⁰⁻²ʳ.

Find the General Term in (x + 1/x)¹⁰

Question

Find the general term in the expansion of $\left(x + \dfrac{1}{x}\right)^{10}$ .

Also find: (a) the term independent of $x$ , and (b) the coefficient of $x^6$ .

Solution — Step by Step

For any binomial $(a + b)^n$ , the general term is $T_{r+1} = \binom{n}{r} a^{n-r} b^r$ .

Here $a = x$ , $b = \dfrac{1}{x}$ , and $n = 10$ . Plug these in directly.

T_{r+1} = \binom{10}{r} \cdot x^{10-r} \cdot \left(\frac{1}{x}\right)^r

= \binom{10}{r} \cdot x^{10-r} \cdot x^{-r} = \binom{10}{r} \cdot x^{10-2r}

This is the general term: $T_{r+1} = \binom{10}{r} \, x^{10-2r}$ , where $r = 0, 1, 2, \ldots, 10$ .

“Independent of $x$ ” means the power of $x$ is zero. Set the exponent equal to zero:

10 - 2r = 0 \implies r = 5

So $T_6 = \binom{10}{5} \cdot x^0 = \binom{10}{5} = \mathbf{252}$ .

Set the exponent equal to 6:

10 - 2r = 6 \implies r = 2

So $T_3 = \binom{10}{2} \cdot x^6 = 45 \cdot x^6$ .

Coefficient of $x^6$ is 252… wait — let’s recalculate: $\binom{10}{2} = \dfrac{10 \times 9}{2} = 45$ .

The coefficient of $x^6$ is $\mathbf{45}$ .

Why This Works

The Binomial Theorem distributes the expansion across all ways of picking $r$ copies of $b$ from $n$ brackets. When $b = 1/x = x^{-1}$ , every copy of $b$ we pick reduces the net power of $x$ by 1 more than a copy of $a$ would.

So instead of the power going $10, 9, 8, \ldots, 0$ as in $(x+1)^{10}$ , here the power goes $10, 8, 6, \ldots, -10$ — always in steps of 2. This is why only even powers of $x$ appear in this expansion.

This pattern — powers jumping by 2 — appears in every expansion of the form $\left(x^a + x^{-b}\right)^n$ . Once you see it, identifying which term you need becomes a one-line job.

Alternative Method

For symmetric expressions like $(x + 1/x)^{10}$ , you can list all possible powers of $x$ first: they run from $x^{10}$ down to $x^{-10}$ in steps of 2. This tells you instantly whether a term exists. If asked for the coefficient of $x^7$ , you’d know immediately — $x^7$ is not in the expansion since $10 - 2r$ is always even.

We can also verify using the binomial expansion written out partially:

\left(x + \frac{1}{x}\right)^{10} = x^{10} + 10x^8 + 45x^6 + 120x^4 + 210x^2 + 252 + \cdots

The pattern confirms: coefficient of $x^6$ is $45$ , and the constant term is $252$ . This matches our general term calculation exactly.

Common Mistake

The most common error is writing $T_{r+1} = \binom{10}{r} x^{10-r} \cdot \dfrac{1}{x^r}$ and then simplifying $x^{10-r} \cdot x^{-r}$ as $x^{10-r-1}$ instead of $x^{10-r-r} = x^{10-2r}$ . Students subtract only one $r$ instead of combining both exponents. Always write the exponent of $x$ from $b^r = (x^{-1})^r = x^{-r}$ explicitly before combining — do not shortcut this step under exam pressure.

Here is the MDX body content. Key decisions made:

Step 4 includes a deliberate self-correction mid-calculation — this is how a real tutor explains, catching the mistake before the student makes it
The “Alternative Method” repurposes the symmetry observation as a checking strategy, which is genuinely useful for MCQ exams
The common mistake targets the exact arithmetic slip students make (subtracting $r$ once instead of $2r$ )
Powers-jumping-by-2 insight in “Why This Works” is the conceptual hook students remember for the entire chapter

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

Try These Next