Examples

Chapter 7 Class 11 Binomial Theorem
Serial order wise

### Transcript

Example 15 Find the term independent of x in the expansion of (āš„ " + " 1/(2 āš„))^18, x > 0. Calculating general term of expansion We know that general term of (a + b)n is Tr+1 = nCr (a)nār . (a)n For general term of expansion (āš„ " + " 1/(2 āš„))^18 Putting n = 18 , a = āš„ , b = 1/(2 āš„) ā“ Tr + 1 = 18Cr (āš„)18 ā r (1/(2 āš„))^š = 18Cr (ćš„")" ć^(1/3 Ć (18 ā š)) (1/2 "Ć" 1/āš„)^š = 18Cr (ćš„")" ć^((18 ā š)/3 ) (1/2)^š (1/āš„)^š = 18Cr (ćš„")" ć^((18 ā š)/3 ) 1/2^š (1/š„^(1/3) )^š = 18Cr (ćš„")" ć^((18 ā š)/3 ) 1/2^š 1/š„^(š/3) = 18Cr (ćš„")" ć^((18 ā š)/3 ) 1/2^š š„^((āš)/3) = 18Cr (ćš„")" ć^((18 ā š)/3 ) š„^((āš)/3) 1/2^š = 18Cr (ćš„")" ć^((18 ā š)/3 ā š/3) 1/2^š = 18Cr (ćš„")" ć^((18 ā š ā š)/3 ) 1/2^š = 18Cr (ćš„")" ć^((18 ā 2š)/3 ) 1/2^š ā“ Tr + 1 = 18Cr (ćš„")" ć^((18 ā 2š)/3 ) 1/2^š We need to find the term independent of x So, power of x is 0 š„^((18 ā2š)/3) = x0 Comparing power (18 ā 2š)/3 = 0 18 ā 2r = 0 18 = 2r 18/2 = r 9 = r r = 9 Putting r = 9 in (1) Tr+1 = 18Cr (ćš„")" ć^((18 ā 2š)/3 ) 1/2^š T9+1 = 18C9 .š„^((18 ā3(9))/3). 1/29 = 18C9 .x0. 1/29 = 18C9 . 1/29 Hence, the term which is independent of x is 10th term = T10 = 18C9 . š/šša