Which term of the expansion contains x^8

Question

Which term of the expansion of (x+5)¹⁰ contains x⁸?

the term number is?

Answer 1

You need to use the binomial expansion, (a+b)ⁿ.  The "k^th" term in an expansion is given by:

 

(n!/(n-k)!k!)*a^n-kb^k

 

Where k = 0, 1, 2, ..., n,  Term numbers, however, are numbered from 1 to n+1 rather than from 0 to n.  Hence the term number, r, is r = k+1  (that is, r = 1, 2, 3, ..., n+1)

 

In your case, (x + 5)¹⁰, a=x, b=5, n=10.  What do you have to set k to to get x^n-k = x⁸?  The term number, then is r = k+1. 

 

Use the expression  (n!/(n-k)!k!)*a^n-kb^k  to determine exactly what the term is.

Answer 2

10(9)(8)(7)=5040

Answer 3

Which term of the expansion of (x+5)^10 contains x^8?

 

The Binomial Theorem says (x+5)^10 equals:

 

C(10,0) x^10 5^0 +

C(10,1) x^9 5^1 +

C(10,2) x^8 5^2 +

...

C(10,8) x^2 5^8

C(10,9) x 5^9
C(10,10) 5^10

 

So x^8 is in the 3rd term if you count from 1, or the 2nd term if you count from 0.

 

(x+5)^10 = (5 + x)^10

 

Which term of the expansion of (5+x)^10 contains x^8?

 

If C(10,3) = 10!/(3!(10-3)!), what does C(10,7) equal?