In each of the following expansions , find the term as stated .
(1) $\displaystyle (1+x)^{10}$ , 5th term
(2) $\displaystyle (2a+b)^{12}$ , 10th term
(3) $\displaystyle (x-\frac{1}{x})^6$ , constant term
Use the formula:
$\displaystyle (a+b)^n=\sum_{k=0}^n {n \choose k} a^{n-k} \cdot b^k$
Keep in mind that you start counting by zero! Thus the 5th term correspond with k = 4.
Therefore the 5th term of $\displaystyle (1+x)^{10} \rightarrow {10\choose 4}1^4\cdot x^6 = 210x^6
$
For the last example: A constant term only occurs if the exponents of x and $\displaystyle \frac1x$ are equal, thus you have to calculate the 4th term with k = 3. Since k is an odd number you get as the constant summand -20.