Skip to content
Exercise 5.1 · Q7

Q.Find the constant term of (2x3−13x2)5\left(2x^3-\dfrac{1}{3x^2}\right)^5.

Puducherry TnboardTextbookSubjectiveImportance★★★★★
7% · 7/95 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Write the general term of (2x3−13x2)5\left(2x^3-\dfrac1{3x^2}\right)^5, find the value of rr that makes the power of xx zero, and evaluate that term.

Step 1. General term. Tr+1=5Cr(2x3)5−r(−13x2)r=5Cr 25−r(−1)r 3−r x3(5−r)−2r=5Cr 25−r(−1)r3−r x15−5rT_{r+1}={}^5C_r(2x^3)^{5-r}\left(-\dfrac1{3x^2}\right)^r={}^5C_r\,2^{5-r}(-1)^r\,3^{-r}\,x^{3(5-r)-2r}={}^5C_r\,2^{5-r}(-1)^r3^{-r}\,x^{15-5r}. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.