find the constant term of the expansion of (3x²+(1/x))^8
I don't know where to start =\
math - Damon, Sunday, February 3, 2008 at 10:03pm
(1/x^8) [ 3 x^10 +1 ]^8
now find the term in x^8 in the expansion of
(3 x^10 + 1)^8
because when you multiply by (1/x^8) it will be constant.
I would use the formula for binomial coefs rather than 9 rows of Pascal
C(n,r) = n!/[r!(n-r)!]
for n = 8 and r = 2
C = 8!/[6!(2!)]
= 8*7/2 = 28
math - Reiny, Monday, February 4, 2008 at 7:46am
The general term in the expansion is
tr+1 = C(8,r)(3x2)8-r(1/x)r
= C(8,r) 3^(8-r) x^(16-3r)
to have a constant term the exponent of x must be zero, or
16-3r = 0
There is not integer solution for this, so the expansion does not have a "constant" term
another way to see it,...
the first term contains x^16, the second term constains x^13, .....
each term will contain an x term