मराठी

Find the expansion of (3x2 – 2ax + 3a2)3 using binomial theorem. - Mathematics

Advertisements
Advertisements

प्रश्न

Find the expansion of (3x2 – 2ax + 3a2)3 using binomial theorem.

बेरीज
Advertisements

उत्तर

Using binomial theorem, the given expression (3x* -2ax +3a* ) can be expanded

[3x2 - a (2x - 3a)]3

= 3C0 (3x2 - 2ax)3 + 3C1(3x2 - 2ax)2 (3a2)+ 3C2(3x2 - 2ax) (3a2)2 + 3C3(3a2)3

= (3x2 - 2ax)3 + 3(9x4 - 12ax3 + 4a2x2)(3a2)+3(3x2 - 2ax)(9a4) + 27a6

= (3x2 - 2ax)3 + 81a2x4 - 108a3x3 + 36a4x2 + 81a4x2 - 54a5x + 27a6

= (3x2 - 2ax)3 + 81a2x4 - 108a3x3 + 117a4x2 - 54a5x + 27a6

Again, by using binomial theorem, we obtain

(3x2 - 2ax)3

= 3C0 (3X2)3 - 3C1 (3X2)2 (2ax) + 3C2 (3X2)(2ax)2 - 3C3 (2ax)3

= 27x6 - 3(9x4) (2ax) + 3 (3x2) (4a2x2) -8a3x3

= 27x6 - 54ax5 + 36a2x4 - 8a3x3

From (1) and (2), we obtain

(3x2 - 2ax + 3a2)3

= 27x6 - 54ax5 + 36a2 x4 - 8a3x3 + 81a2x4 - 108a3x3  + 117a4 x2 - 54a5x + 27a6

= 27x6 - 54ax5 + 117a2 x4 - 116a3 x3 + 117a4 x2 - 54a5x + 27a6

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 8: Binomial Theorem - Miscellaneous Exercise [पृष्ठ १७६]

APPEARS IN

एनसीईआरटी Mathematics [English] Class 11
पाठ 8 Binomial Theorem
Miscellaneous Exercise | Q 10 | पृष्ठ १७६

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Expand the expression (1– 2x)5


Expand the expression: (2x – 3)6


Using Binomial Theorem, evaluate the following:

(96)3


Using binomial theorem, evaluate f the following:

(101)4


Using Binomial Theorem, indicate which number is larger (1.1)10000 or 1000.


Prove that `sum_(r-0)^n 3^r  ""^nC_r = 4^n`


Find ab and n in the expansion of (a + b)n if the first three terms of the expansion are 729, 7290 and 30375, respectively.


Find the value of `(a^2 + sqrt(a^2 - 1))^4 + (a^2 - sqrt(a^2 -1))^4`


If n is a positive integer, prove that \[3^{3n} - 26n - 1\]  is divisible by 676.

 
 

Using binomial theorem determine which number is larger (1.2)4000 or 800?

 

Find the value of (1.01)10 + (1 − 0.01)10 correct to 7 places of decimal.

 

Show that  \[2^{4n + 4} - 15n - 16\]  , where n ∈  \[\mathbb{N}\]  is divisible by 225.

 
  
  

Find the 4th term from the end in the expansion of `(x^3/2 - 2/x^2)^9`


Find the coefficient of x11 in the expansion of `(x^3 - 2/x^2)^12`


If n is a positive integer, find the coefficient of x–1 in the expansion of `(1 + x)^2 (1 + 1/x)^n`


Which of the following is larger? 9950 + 10050  or 10150


Find the coefficient of x50 after simplifying and collecting the like terms in the expansion of (1 + x)1000 + x(1 + x)999 + x2(1 + x)998 + ... + x1000 .


The total number of terms in the expansion of (x + a)51 – (x – a)51 after simplification is ______.


If the coefficients of x7 and x8 in `2 + x^n/3` are equal, then n is ______.


If (1 – x + x2)n = a0 + a1 x + a2 x2 + ... + a2n x2n , then a0 + a2 + a4 + ... + a2n equals ______.


The number of terms in the expansion of (a + b + c)n, where n ∈ N is ______.


Find the coefficient of x in the expansion of (1 – 3x + 7x2)(1 – x)16.


Find the coefficient of x15 in the expansion of (x – x2)10.


Find the coefficient of x4 in the expansion of (1 + x + x2 + x3)11.


The two successive terms in the expansion of (1 + x)24 whose coefficients are in the ratio 1:4 are ______.


The coefficient of a–6b4 in the expansion of `(1/a - (2b)/3)^10` is ______.


Number of terms in the expansion of (a + b)n where n ∈ N is one less than the power n.


If the coefficients of (2r + 4)th, (r – 2)th terms in the expansion of (1 + x)18 are equal, then r is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×