# Thread:

1. ## Binomial expansion

In each of the following expansions , find the term as stated .

(1) $(1+x)^{10}$ , 5th term

(2) $(2a+b)^{12}$ , 10th term

(3) $(x-\frac{1}{x})^6$ , constant term

2. Originally Posted by thereddevils
In each of the following expansions , find the term as stated .

(1) $(1+x)^{10}$ , 5th term

(2) $(2a+b)^{12}$ , 10th term

(3) $(x-\frac{1}{x})^6$ , constant term
Use the formula:

$(a+b)^n=\sum_{k=0}^n {n \choose k} a^{n-k} \cdot b^k$

Keep in mind that you start counting by zero! Thus the 5th term correspond with k = 4.

Therefore the 5th term of $(1+x)^{10} \rightarrow {10\choose 4}1^4\cdot x^6 = 210x^6<br />$

For the last example: A constant term only occurs if the exponents of x and $\frac1x$ 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.

3. Originally Posted by earboth
Therefore the 5th term of $(1+x)^{10} \rightarrow {10\choose 4}1^4\cdot x^6 = 210x^6<br />$

I guess there is a slight typo , should be 210x^4

4. Originally Posted by mathaddict
I guess there is a slight typo , should be 210x^4
With $(1+x)^{10}$ the 5th term is:

${10\choose 4}1^4\cdot x^6 = 210x^6$ and the 7th term is ${10\choose 6}1^6\cdot x^4 = 210x^4<br />$

Mostly the binoms are given as $(x+1)^{10}$ and then you would have been right.
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