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Binomial Expansion - Finding the term independent of n ...


Apr 8, 2016 ... For a general n, the constant term in f(x)=(1+x/5)n(2−3/x)2. can be found by observing first that (2−3/x)2=2−12x−1+9x−2. Then we see that the ...


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Binomial Theorem Find Term independent of variable x - YouTube


Oct 4, 2017 ... Related IB Math Test: https://www.youtube.com/watch?v=ChHhF1GvZzY&index= 9&list=PLJ-ma5dJyAqqN8RzW7LQ7M7lRUPsHSDoP.


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Expanding binomials (video) | Polynomials | Khan Academy


Sal expands (3y^2+6x^3)^5 using the binomial theorem and Pascal's ... CCSS Math: ... (ii) With this value of n, find the term independent of x in the expansion ...


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BINOMIAL THEOREM


In the expansion, the first term is raised to the power of the binomial and in each subsequent terms the ... 130 EXEMPLAR PROBLEMS – MATHEMATICS. Index of ... Example 7 Find the term independent of x in the expansion of. 10. 2. 3. 23.


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Binomial Expansions Examples (examples, solutions, worksheets ...


Examples, solutions, videos, activities, and worksheets that are suitable for A Level Maths. Terms Independent of x or y in the Binomial Expansion Finding the term ...


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Binomial Theorem - Properties, Terms in Binomial Expansion ...


The term Independent of in the expansion of [axp + (b/xq)]n is ... Binomial theorem has a wide range of application in mathematics like finding the remainder, ...


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Integral $\int_{-1}^1\frac1x\sqrt{\frac{1+x}{1-x}}


Mathematics Stack Exchange ... First sub t=(1x)/(1+x), dt=−2/(1+x)2dx to get ... [2 log(1+u1−u)log(5−2u2+u41−2u2+5u4)]10−32∫10du(u5−6u3+u)(u4−2u2+5)(5u 4−2u2+1)log(1+u1−u) ... Expanding the square, we get 3 terms: ... This leaves the second term. ... In the second step we show that I(r,s) is independent of s.


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Error function - Wikipedia


Name[edit]. The name "error function" and its abbreviation erf were proposed by J. W. L. ... In mathematics, the error function (also called the Gauss error function), often denoted by erf, ... In statistics, for nonnegative values of x, the error function has the following ... The error function at +∞ is exactly 1 (see Gaussian integral).


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Math Camp


1) Properties of Expectations, Changing Coordinates; 2) Moment Generating Functions. Two random variables X and Y are independent if for any two sets of ...


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Krishna
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Given expansion

[math] (\sqrt[3]{x} + \frac{1}{2\sqrt[3]{x}})^{18} [/math], x > 0


Find the term independent


We know that

The (r + 1)th term of the expansion (x + y)^n is given by T_{r + 1} = ^nC_r x^{n-r} y^r


[math] (\sqrt[3]{x} + \frac{1}{2\sqrt[3]{x}})^{18} [/math]


  n = 18  [math]x\ =\ \sqrt[3]{x}\ \ and\ \ \ y\ =\ \frac{1}{2\sqrt[3]{x}}[/math]


[math] T_{r + 1} = ^{18}C_r (\sqrt[3]{x})^{18 - r} (\frac{1}{2\sqrt[3]{x}})^r [/math]


T_{r+1}=^{18}C_r(x^{\frac{1}{3}})^{18-r}\ \left(\frac{1}{2}\right)^{^r}\ \frac{1}{\left(x^{\frac{1}{3}}\right)^{^r}}


T_{r+1}=^{18}C_r\ \ \ x^{\frac{18\ -\ r}{3}}\ \ \ x^{-\frac{\left\{r\right\}}{3}}\left(\frac{1}{2}\right)^{^r}\ ^{ }


T_{r+1}=^{18}C_r\ \ \ \ \ x^{\frac{18\ -\ r}{3}\ -\ \frac{r}{3}}\ \ \left(\frac{1}{2}\right)^{^r}


T_{r+1}=^{18}C_r\ \ \ \ \ x^{\frac{18\ -\ r-r}{3}\ }\ \ \left(\frac{1}{2}\right)^r


T_{r+1}=^{18}C_r\ \ \ \ \ x^{\frac{18\ -2\ r}{3}\ }\ \ \left(\frac{1}{2}\right)^r.................(1)


Since we have to find a term independent of x, i.e., term not having x,

so take    x^{ \frac{18 - 2r}{3}} = x^{0}


               \frac{18 - 2r}{3} = 0


                 18 - 2r = 0


                 18 = 2r


                 r = \frac{18}{2}


                 r = 9


Substituting the r = 9 in equation (1)

  

            T_{r+1}=^{18}C_r\ \ \ \ \ x^{\frac{18\ -2\ r}{3}\ }\ \ \left(\frac{1}{2}\right)^r


            T_{r+1}=^{18}C_9\ \ \ \ \ x^{\frac{18\ -2\ \left(9\right)}{3}\ }\ \ \left(\frac{1}{2}\right)^9


          T_{r+1}=^{18}C_9\ \ \ \left(\frac{1}{2}\right)^9\ x^0


Hence, required term = ^{18}C_{9\ }\ \frac{1}{2^9}