Answers>Maths>IB>Article

Prove by mathematical induction that (2C2)+(3C2)+(4C2)+...+(n-1C2) = (nC3).

Firstly, show the equation is true for n = 3 (as this is the samllest n that nC3 is defined): LHS = (2C2) = 1 = (3C3) = RHStherefore, true for n=3.
Then assume true for n = k:(2C2)+(3C2)+(4C2)+...+(k-1C2) = (kC3).
Concider n = k-1:(2C2)+(3C2)+(4C2)+...+(k-1C2)+(kC2) = (kC3)+(kC2) = [k!/(k-3)!3!] + [k!/(k-2)!2!] = (k!/3!)[(1/(k-3)!)+3/(k-2)!] = (k!/3!)[(k-2+3)/(k-2)!] = (k!/3!)[(k+1)/(k-2)!] = [(k+1)!/3!(k-2)!] = (k+1)C3
Equation is true for n = 3. If true for n = k, it is true for n = k+1. Therefore the equation is true for all n >= 3 by induction.

HX
Answered by Henry X. Maths tutor

14693 Views

See similar Maths IB tutors

Related Maths IB answers

All answers ▸

Solve x^2 + 2x = 48 for all values of x


How to integrate ∫〖3x/√(1-x^2 ) dx〗?


How to find the derivative of sqrt(x) from first principles?


Let f (x) = sin(x-1) , 0 ≤ x ≤ 2 π + 1 , Find the volume of the solid formed when the region bounded by y =ƒ( x) , and the lines x = 0 , y = 0 and y = 1 is rotated by 2π about the y-axis.


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences