Binomial expansion with different binomials

**grillage** · October 2nd 2012, 01:57 AM

I have a set of questions like the following:

For the function $\left ( 1+x \right )\left ( 1-x \right )^{9}$ , find the term containing $x^{r}$ and hence express the expansion in sigma notation.

The answer for this particular problem is given at the back of the book as

$\sum_{r=0}^{9}\left ( -1 \right )^{r}\frac{9!x^{r}\left ( 10-2r \right )}{r!\left ( 10-r \right )!}$

I get that there is some manipulation of the binomial coefficient, but I am wondering if there is a standard approach to these types of problems, and it confuses me that r still goes to 9 when there should now be 11 terms it seems.

Thoughts appreciated.

Andrew

**johnsomeone** · October 2nd 2012, 07:39 AM

Here's the direct approach - it's just some common series index manipulations:

I'll write $C(n, a) = \frac{n!}{a!(n-a)!}$ for "n choose a".

$(1-x)^9 = \sum_{i=0}^9 (-1)^i C(9, i) x^i$ , so

$(1+x)(1-x)^9 = (1+x)\sum_{i=0}^9 (-1)^i C(9, i) x^i = \sum_{i=0}^9 (-1)^i C(9, i) x^i + x\sum_{i=0}^9 (-1)^i C(9, i) x^i$

$= \sum_{i=0}^9 (-1)^i C(9, i) x^i + \sum_{i=0}^9 (-1)^i C(9, i) x^{i+1}$ (now substitute r = i+1 in the 2nd sum)

$= \sum_{i=0}^9 (-1)^i C(9, i) x^i + \sum_{r=1}^{10} (-1)^{r-1} C(9, r-1) x^r$ (now peel off the r=10)

$= \sum_{i=0}^9 (-1)^i C(9, i) x^i + \sum_{r=1}^{9} (-1)^{r-1} C(9, r-1) x^r + (-1)^{10-1} C(9, 10-1) x^{10}$ (note: C(9, 9) = 1)

$= \sum_{i=0}^9 (-1)^i C(9, i) x^i + \sum_{r=1}^{9} (-1)^{r-1} C(9, r-1) x^r - x^{10}$ (now peel off the i=0, set r = i)

$= \sum_{r=1}^9 (-1)^r C(9, r) x^r + (-1)^0 C(9, 0) x^0 - \sum_{r=1}^{9} (-1)^r C(9, r-1) x^r - x^{10}$ (note signs in 2nd sum)

$= \sum_{r=1}^9 (-1)^r C(9, r) x^r + 1 - \sum_{r=1}^{9} (-1)^r C(9, r-1) x^r - x^{10}$ (note: C(9,0) = 1)

$= 1 - x^{10} + \sum_{r=1}^9 (-1)^r C(9, r) x^r - \sum_{r=1}^{9} (-1)^r C(9, r-1) x^r$ (now combine the sums)

$= 1 - x^{10} + \sum_{r=1}^9 (-1)^r \left( C(9, r) - C(9, r-1) \right) x^r$ .

Thus $(1+x)(1-x)^9 = 1 - x^{10} + \sum_{r=1}^9 (-1)^r \left( C(9, r) - C(9, r-1) \right) x^r$ .

Now simplify $C(9, r) - C(9, r-1)$

$= \frac{9!}{r!(9-r)!} - \frac{9!}{(r-1)!(9-(r-1))!} = \frac{9!}{r!(9-r)!} - \frac{9!}{(r-1)!(10-r)!}$

$= \frac{9!}{(r-1)!(9-r)!} \left( \frac{1}{r} - \frac{1}{10-r} \right) = \frac{9!}{(r-1)!(9-r)!} \left( \frac{10-r}{r(10-r)} - \frac{r}{r(10-r)} \right)$

$= \frac{9!}{(r-1)!(9-r)!} \left( \frac{10-2r}{r(10-r)} \right) = \frac{9!(10-2r)}{r!(10-r)!}$ .

Thus $(1+x)(1-x)^9 = 1 - x^{10} + \sum_{r=1}^9 (-1)^r \frac{9!(10-2r)}{r!(10-r)!} x^r$ .

Note that evaluating at $r=0: (-1)^0 \frac{9!(10-2(0))}{0!(10-0)!} x^0 = (1)\frac{9!(10)}{10!}(1) = 1$ .

Thus $(1+x)(1-x)^9 = - x^{10} + \sum_{r=0}^9 (-1)^r \frac{9!(10-2r)}{r!(10-r)!} x^r$ (treat the leading 1 as r=0).

****
Note that your book was wrong to exclude the $- x^{10}$ , since $(1+x)(1-x)^9$ is a 10th degree polynomial, but the answer your book gave was only a 9th degree polynomial.
****

**grillage** · October 2nd 2012, 08:40 AM

Thanks for the response, I will take a closer look this evening when I have more time. It seems odd however that this textbook should include an example of this nature which seems far more challenging than the level it is intended for. And the omission of the 11th term did confuse me as I said at the beginning.

Andrew

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