# Newton binomial

Mary
Visitor
2016-06-27 06:08:28
#1
Hi,
I need help with Newtons' binomial. If we developing the binomial $left&space;(&space;xsqrt[4]{x}+frac{1}{sqrt{x}}&space;right&space;)^{n}$ the sum of rank 2k coefficients is 128. Which is the term which contains x3?

## RE: Newton binomial

Mary
Visitor
2016-06-27 06:11:16
#2
ooofff!!!!
this is a problem with equations editor in this forum:
the binomial is
$\left ( x\sqrt[4]{x}+\frac{1}{\sqrt{x}} \right )^{n}$

## RE: Newton binomial general term

TeacherM
Visitor
2016-06-27 06:39:28
#3
We have the sum of even coefficients:
$\mathbf{C}_{n}^{0}+\mathbf{C}_{n}^{2}+\mathbf{C}_{n}^{4}+\mathbf{C}_{n}^{6}+...=2^{n-1}=128 => n=8$
And the formula in general term (in Newtons' binomial) is:
$T_{k+1}=\mathbf{C}_{8}^{k}\cdot (x\sqrt[4]{x})^{8-k}\cdot \left ( \frac{1}{\sqrt{x}} \right )^{k}=\mathbf{C}_{8}^{k}(x^{\frac{5}{4}})^{8-k}x^{\frac{-k}{2}}=\mathbf{C}_{8}^{k}x\frac{40-7k}{4}$
so though, $\frac{40-7k}{4}=3=>k=4$

That's means that the term k+1 = 5,i.e. fifth term which contains   x3

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