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

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- Feb 15th 2009, 12:49 AMthereddevilsBinomial expansion
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 - Feb 15th 2009, 01:32 AMearboth
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. - Feb 15th 2009, 02:38 AMmathaddict
- Feb 15th 2009, 04:22 AMearboth
With $\displaystyle (1+x)^{10}$ the 5th term is:

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

$

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