Find the coefficient of 1x17 in the expansion of (x4-1x3)15 - Mathematics

Advertisement Remove all ads
Advertisement Remove all ads
Advertisement Remove all ads
Sum

Find the coefficient of `1/x^17` in the expansion of `(x^4 - 1/x^3)^15`

Advertisement Remove all ads

Solution

The given expression is `(x^4 - 1/x^3)^15`

General Term `"T"_(r + 1) = ""^n"C"_r x^(n - r) y^r`

= `""^15"C"_r  (x^4)^(15 - r) (- 1/x^3)^r`

= `""^15"C"_r  (x)^(60 - 4r) (-1)^r * 1/x^(3r)`

= `""^15"C"_r (-1)^r * 1/(x^(3r - 60 + 4r))`

= `""^15"C"_r (-1)^r * 1/(x^(7r - 60))`

To find the coefficient of `1/x^17`

Put 7r – 60 = 17

⇒ 7r = 60 + 17

⇒ 7r = 77

∴ r = 11

Putting the value of r in the above expression, we get

= `""^15"C"_11 (-1)^11 * 1/x^17`

= `- ""^15"C"_4 * 1/x^17`

= `- (15 xx 14 xx 13 xx 12)/(4 xx 3 xx 2 xx 1) * 1/x^17`

= `- 1365 * 1/x^17`

Hence, the coefficient of `1/x^17` = – 1365

Concept: General and Middle Terms
  Is there an error in this question or solution?
Chapter 8: Binomial Theorem - Exercise [Page 143]

APPEARS IN

NCERT Mathematics Exemplar Class 11
Chapter 8 Binomial Theorem
Exercise | Q 7 | Page 143

Video TutorialsVIEW ALL [1]

Share
Notifications

View all notifications


      Forgot password?
View in app×