English

Write the General Term in the Expansion of (X2 – Yx)12, X ≠ 0

Advertisements
Advertisements

Question

Write the general term in the expansion of (x2 – yx)12x ≠ 0

Advertisements

Solution

shaalaa.com
  Is there an error in this question or solution?

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

Find the coefficient of x5 in (x + 3)8


Write the general term in the expansion of (x2 – y)6


Find the middle terms in the expansions of `(x/3 + 9y)^10`


In the expansion of (1 + a)m + n, prove that coefficients of am and an are equal.


Find a positive value of m for which the coefficient of x2 in the expansion

(1 + x)m is 6


Find the middle term in the expansion of: 

(ii)  \[\left( \frac{a}{x} + bx \right)^{12}\]

 


Find the middle terms(s) in the expansion of: 

(i) \[\left( x - \frac{1}{x} \right)^{10}\]

 


Find the middle terms(s) in the expansion of:

(ii)  \[\left( 1 - 2x + x^2 \right)^n\]


Find the middle terms(s) in the expansion of: 

(vi)  \[\left( \frac{x}{3} + 9y \right)^{10}\]

 


Find the middle terms(s) in the expansion of: 

(vii) \[\left( 3 - \frac{x^3}{6} \right)^7\]

  


Find the middle terms(s) in the expansion of:

(x)  \[\left( \frac{x}{a} - \frac{a}{x} \right)^{10}\]

 


Find the term independent of x in the expansion of the expression: 

(iii)  \[\left( 2 x^2 - \frac{3}{x^3} \right)^{25}\]

 


Find the term independent of x in the expansion of the expression: 

(v)  \[\left( \frac{\sqrt{x}}{3} + \frac{3}{2 x^2} \right)^{10}\]

 


Find the term independent of x in the expansion of the expression: 

(x) \[\left( \frac{3}{2} x^2 - \frac{1}{3x} \right)^6\]

 


The coefficients of 5th, 6th and 7th terms in the expansion of (1 + x)n are in A.P., find n.

 

If the coefficients of 2nd, 3rd and 4th terms in the expansion of (1 + x)2n are in A.P., show that  \[2 n^2 - 9n + 7 = 0\]

 


If in the expansion of (1 + x)n, the coefficients of pth and qth terms are equal, prove that p + q = n + 2, where  \[p \neq q\]

 


Write the middle term in the expansion of  \[\left( x + \frac{1}{x} \right)^{10}\]

 

Find the sum of the coefficients of two middle terms in the binomial expansion of  \[\left( 1 + x \right)^{2n - 1}\]

 

If A and B are the sums of odd and even terms respectively in the expansion of (x + a)n, then (x + a)2n − (x − a)2n is equal to


The number of irrational terms in the expansion of \[\left( 4^{1/5} + 7^{1/10} \right)^{45}\]  is

 

If an the expansion of \[\left( 1 + x \right)^{15}\]   , the coefficients of \[\left( 2r + 3 \right)^{th}\text{  and  } \left( r - 1 \right)^{th}\]  terms are equal, then the value of r is

 

In the expansion of \[\left( x - \frac{1}{3 x^2} \right)^9\]  , the term independent of x is

 

If in the expansion of (1 + y)n, the coefficients of 5th, 6th and 7th terms are in A.P., then nis equal to


In the expansion of \[\left( \frac{1}{2} x^{1/3} + x^{- 1/5} \right)^8\] , the term independent of x is

 

Find the term independent of x, x ≠ 0, in the expansion of `((3x^2)/2 - 1/(3x))^15`


Show that the middle term in the expansion of `(x - 1/x)^(2x)` is `(1 xx 3 xx 5 xx ... (2n - 1))/(n!) xx (-2)^n`


Find the term independent of x in the expansion of (1 + x + 2x3) `(3/2 x^2 - 1/(3x))^9`


Middle term in the expansion of (a3 + ba)28 is ______.


The position of the term independent of x in the expansion of `(sqrt(x/3) + 3/(2x^2))^10` is ______.


The number of terms in the expansion of [(2x + y3)4]7 is 8.


The sum of coefficients of the two middle terms in the expansion of (1 + x)2n–1 is equal to 2n–1Cn


The last two digits of the numbers 3400 are 01.


If the 4th term in the expansion of `(ax + 1/x)^n` is `5/2` then the values of a and n respectively are ______.


The middle term in the expansion of (1 – 3x + 3x2 – x3)6 is ______.


Let for the 9th term in the binomial expansion of (3 + 6x)n, in the increasing powers of 6x, to be the greatest for x = `3/2`, the least value of n is n0. If k is the ratio of the coefficient of x6 to the coefficient of x3, then k + n0 is equal to ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×