SOLUTION: For the expansion of (k+t)^22 ,state a. the number of terms b. the degree of each term c. the first 4 terms in the expansion, without coefficients d. the coefficients of the f

Algebra ->  Permutations -> SOLUTION: For the expansion of (k+t)^22 ,state a. the number of terms b. the degree of each term c. the first 4 terms in the expansion, without coefficients d. the coefficients of the f      Log On


   

Question 249859: For the expansion of (k+t)^22 ,state
a. the number of terms
b. the degree of each term
c. the first 4 terms in the expansion, without coefficients
d. the coefficients of the first three terms
My answers
a. 23
b. 22
c. k^22, k^21t, k^20t^2, and k^19t^3
d. 1, 22 , not sure what the 3rd term is???

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!

The binomial expansion is easier if you don't call the first
term "the first term" but instead call it "the zero-th term",
and call the second term "the first term" and then call the last
term the nth term, (which is really the (n+1)st term).

If you use that "start counting from 0 instead of 1" convention 
then the Rth term of %28K%2BT%29%5EN

is NCR which is sometimes written %22C%28N%2CR%29%22 and sometimes
written %28matrix%282%2C1%2CN%2CR%29%29 but it always means N%21%2F%28R%21%28N-R%29%21%29%7D%7D%29.%0D%0A%0D%0AIn+%7B%7B%7B%28k%2Bt%29%5E22, 

the 0th term's coefficient is 22C0 or %22C%2822%2C0%29%22 or %28matrix%282%2C1%2C22%2C0%29%29

which means 22%21%2F%280%21%2822-0%29%21%29=+22%21%2F%281%2A22%21%29=22%21%2F22%21+=+1

The 1st term's coefficient is 22C1 or %22C%2822%2C1%29%22 or %28matrix%282%2C1%2C22%2C1%29%29

which means 22%21%2F%281%21%2822-1%29%21%29=+22%21%2F%281%2A21%21%29=22%21%2F21%21 

=  =

=  =

22

The 2nd term's coefficient is 22C2 or %22C%2822%2C2%29%22 or %28matrix%282%2C1%2C22%2C2%29%29

which means 22%21%2F%282%21%2822-2%29%21%29=+22%21%2F%282%21%2A20%21%29=22%21%2F%282%2120%21%29 

=  =

=  =

%2822%2A21%29%2F%282%2A1%29+=+231

You can also just skip all that cancelling and observe
the pattern of what's always left after cancelling:

If you needed the 3rd term's coefficient (really the 4th term's), 
it would be

%2822%2A21%2A20%29%2F%283%2A2%2A1%29 

If you needed the 4th term's coefficient (really the 5th term), it would be

%2822%2A21%2A20%2A19%29%2F%284%2A3%2A2%2A1%29 

If you needed the 5th term's coefficient (really the 6th term), it would be

%2822%2A21%2A20%2A19%2A18%29%2F%285%2A4%2A3%2A2%2A1%29

etc., etc.

So to get the numerator of any term's coefficient, you start with
the outer exponent, and multiply it by 1 less, then 1 less, etc.
until you have as many factors as the number of term (starting
counting from zero, not 1). Then the denominator is just that
number factorial.  There are always as many factors on top
as the number of term (starting counting form 0, not 1). And
of course the denominator has that same number of factors, too.

Edwin