मराठी

Using Binomial Theorem, Write Down the Expansions . (Iii) ( X − 1 X ) 6 - Mathematics

Advertisements
Advertisements

प्रश्न

Using binomial theorem, write down the expansions  .

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

Advertisements

उत्तर

(iii) \[\left( x - \frac{1}{x} \right)^6 \]
\[ = ^{6}{}{C}_0 x^6 \left( \frac{1}{x} \right)^0 - ^{6}{}{C}_1 x^5 \left( \frac{1}{x} \right)^1 +^{6}{}{C}_2 x^4 \left( \frac{1}{x} \right)^2 - ^{6}{}{C}_3 x^3 \left( \frac{1}{x} \right)^3 + ^{6}{}{C}_4 x^2 \left( \frac{1}{x} \right)^4 -^6 C_5 x^1 \left( \frac{1}{x} \right)^5 + ^{6}{}{C}_6 x^0 \left( \frac{1}{x} \right)^6 \]
\[ = x^6 - 6 x^5 \times \frac{1}{x} + 15 x^4 \times \frac{1}{x^2} - 20 x^3 \times \frac{1}{x^3} + 15 x^2 \times \frac{1}{x^4} - 6 x \times \frac{1}{x^5} + \frac{1}{x^6}\]
\[ = x^6 - 6 x^4 + 15 x^2 - 20 + \frac{15}{x^2} - \frac{6}{x^4} + \frac{1}{x^6}\]

shaalaa.com
Introduction of Binomial Theorem
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 18: Binomial Theorem - Exercise 18.1 [पृष्ठ ११]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
पाठ 18 Binomial Theorem
Exercise 18.1 | Q 1.03 | पृष्ठ ११

संबंधित प्रश्‍न

Using binomial theorem, write down the expansions  :

(iv)  \[\left( 1 - 3x \right)^7\]

 


Using binomial theorem, write down the expansions  :

(vii)  \[\left( \sqrt[3]{x} - \sqrt[3]{a} \right)^6\]

 


Using binomial theorem, write down the expansions  :

(viii)  \[\left( 1 + 2x - 3 x^2 \right)^5\]

 


Using binomial theorem, write down the expansions  :

(ix) \[\left( x + 1 - \frac{1}{x} \right)\]

 


Evaluate the 

(i)\[\left( \sqrt{x + 1} + \sqrt{x - 1} \right)^6 + \left( \sqrt{x + 1} - \sqrt{x - 1} \right)^6\]

 


Evaluate the

(vi)  \[\left( 2 + \sqrt{3} \right)^7 + \left( 2 - \sqrt{3} \right)^7\]


Evaluate the

(vii) \[\left( \sqrt{3} + 1 \right)^5 - \left( \sqrt{3} - 1 \right)^5\]

 


Evaluate the

(viii)  \[\left( 0 . 99 \right)^5 + \left( 1 . 01 \right)^5\]

 

Evaluate the

(ix) \[\left( \sqrt{3} + \sqrt{2} \right)^6 - \left( \sqrt{3} - \sqrt{2} \right)^6\]

 


Evaluate the

(x) \[\left\{ a^2 + \sqrt{a^2 - 1} \right\}^4 + \left\{ a^2 - \sqrt{a^2 - 1} \right\}^4\]

 

Find \[\left( x + 1 \right)^6 + \left( x - 1 \right)^6\] . Hence, or otherwise evaluate \[\left( \sqrt{2} + 1 \right)^6 + \sqrt{2} - 1^6\] .

 

 


Using binomial theorem evaluate .

(iii) (101)4

 


Using binomial theorem evaluate .

(iv) (98)5

 

Using binomial theorem, prove that  \[3^{2n + 2} - 8n - 9\]  is divisible by 64, \[n \in N\] .

 

Find the coefficient of: 

(iii)  \[x^{- 15}\]  in the expansion of  \[\left( 3 x^2 - \frac{a}{3 x^3} \right)^{10}\]

 

 


Find the coefficient of: 

(iv)  \[x^9\]  in the expansion of  \[\left( x^2 - \frac{1}{3x} \right)^9\]

 

 


Find the coefficient of: 

(v)  \[x^m\]  in the expansion of  \[\left( x + \frac{1}{x} \right)^n\]

 

 


Find the coefficient of: 

(vi) x in the expansion of  \[\left( 1 - 2 x^3 + 3 x^5 \right) \left( 1 + \frac{1}{x} \right)^8\]

 

Find the coefficient of: 

(vii) \[a^5 b^7\]  in the expansion of  \[\left( a - 2b \right)^{12}\]

 
 

Find the coefficient of: 

(viii) x in the expansion of \[\left( 1 - 3x + 7 x^2 \right) \left( 1 - x \right)^{16}\]

 

Write the sum of the coefficients in the expansion of \[\left( 1 - 3x + x^2 \right)^{111}\]

 

If a and b denote respectively the coefficients of xm and xn in the expansion of \[\left( 1 + x \right)^{m + n}\], then write the relation between a and b.

 
 

If a and b denote the sum of the coefficients in the expansions of \[\left( 1 - 3x + 10 x^2 \right)^n\]  and \[\left( 1 + x^2 \right)^n\]  respectively, then write the relation between a and b.

 
 
 

If the coefficient of x in \[\left( x^2 + \frac{\lambda}{x} \right)^5\]  is 270, then \[\lambda =\]

 
 

The coefficient of x4 in \[\left( \frac{x}{2} - \frac{3}{x^2} \right)^{10}\] is

 

If  \[T_2 / T_3\]  in the expansion of \[\left( a + b \right)^n \text{ and } T_3 / T_4\]  in the expansion of \[\left( a + b \right)^{n + 3}\]  are equal, then n =

 
 

If the sum of the binomial coefficients of the expansion \[\left( 2x + \frac{1}{x} \right)^n\]  is equal to 256, then the term independent of x is

  

The coefficient of x5 in the expansion of \[\left( 1 + x \right)^{21} + \left( 1 + x \right)^{22} + . . . + \left( 1 + x \right)^{30}\]

 

The coefficient of x8 y10 in the expansion of (x + y)18 is


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×