Binomial: The Term Independent of x

“The term independent of $x$” is the constant term, the one with $x^0$. Using the general term, write the power of $x$ as an expression in $r$, set it to zero, and solve for $r$. Then compute that one term.

Find the term independent of $x$ in the expansion of $\left(2x + \dfrac{1}{x}\right)^6$. [5]

The working

Step 1, write the general term for $\left(2x + \frac{1}{x}\right)^6$, with $a = 2x$, $b = \frac{1}{x}$: $\binom{6}{r}(2x)^{6-r}\left(\frac{1}{x}\right)^r = \binom{6}{r}2^{6-r}\,x^{6-r}\,x^{-r} = \binom{6}{r}2^{6-r}\,x^{6-2r} \quad \text{(M1)}$

Step 2, set the power of $x$ to zero (independent of $x$): $6 - 2r = 0 \;\Rightarrow\; r = 3 \quad \text{(M1, A1)}$

Step 3, compute the term at $r = 3$: $\binom{6}{3}2^{6-3} = 20 \times 2^3 = 20 \times 8 = 160 \quad \text{(M1, A1)}$

So the term independent of $x$ is $160$.

Where the marks are won and lost

• Combine the powers of $x$ correctly: $x^{6-r} \cdot x^{-r} = x^{6-2r}$. Getting this exponent right is what lets you find $r$.
• Set the power to zero ($6 - 2r = 0$), giving $r = 3$. Setting it to another value answers a different question.
• Include the $2^{6-r}$ factor: $\binom{6}{3} = 20$ and $2^3 = 8$ give $160$. Forgetting the $2^3$ is the common error.

Common mistakes

• Writing $x^{6-r} \cdot x^{-r}$ as $x^{6}$ or $x^{6-r-1}$ (mishandling the negative power).
• Setting $6 - 2r$ to $1$ or another value instead of $0$.
• Dropping the $2^{6-r}$ coefficient.

Full method: Binomial Expansion notes. See also Binomial Coefficient of a Term. Topic home: Series pillar.