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October 22, 2014

October 22, 2014

Posted by **Liz** on Sunday, February 3, 2008 at 9:35pm.

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 7:46amThe general term in the expansion is

t_{r+1}= C(8,r)(3x^{2})^{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

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