# Write the series in expanded form. 5 ∑1/2k K=1 Write the first 3 terms of the expansion of (a-b)^6. What are the last 3 terms of (a-b)^7? Show work.

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### 1 Answer

1. Write `sum_(k=1)^5 1/2 k` in expanded form:

k is the index. k=1 is the lower bound, k=5 the upper bound so there are 5 terms in the series expansion. To find each term, substitute the value of k into the expression:

k=1: 1/2(1)=1/2

k=2: 1/2(2)=1 etc...

**The expanded series is 1/2+1+3/2+2+5/2**

2. Write the first three terms of `(a-b)^6` :

The binomial expansion is given by:

`(a-b)^n=C(n,0)a^nb^0+C(n,1)a^(n-1)b^1+ +c(n,n-1)a^1b^(n-1)+C(n,n)a^0b^n`

Here C(n,k) is the number of combinations possible if you choose k items from n items. (This is also written as `_n C _k` )

So the first three terms are:

`C(6,0)a^6b^0-C(6,1)a^5b^1+C(6,2)a^5b^2`

** Note the alternating sign -- the term will be positive if the exponent on b is even, negative if the exponent is odd.)

**Then the first three terms are `a^6-6a^5b+15a^4b^2` **

** To calculate C(n,k) you can use a calculator, or the formula :

`C(n,k)=(n!)/(k!(n-k)!)` where n! is the factorial function.

You could also get the coefficients from Pascal's triangle.

3. The last three terms of `(a-b)^7` will be:

`-C(7,5)a^2b^5+C(7,6)a^1b^6-C(7,7)a^0b^7` or

`-21a^2b^5+7ab^6-b^7`