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.
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=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:
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:
** 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: