Question

1÷1.2.3+1÷2.3.4+1÷3.4.5+....+1÷n(n+1)(n+2)=n(n+3)÷4(n+1)(n+2) solve by mathematical induction

Answer 1

1÷1.2.3+1÷2.3.4+1÷3.4.5+....+1÷n(n+1)(n+2)=n(n+3)÷4(n+1)(n+2) solve by mathematical induction

Solution,

Let suppose,

P(n):1÷1.2.3+1÷2.3.4+1÷3.4.5+....+1÷n(n+1)(n+2)=n(n+3)÷4(n+1)(n+2)

and put n=1 (on assumption basis)

L.H.S=1/1.2.3=1/6

R.H.S=n(n+3)÷4(n+1)(n+2)=1(1+3)÷4(1+1)(1+2)

R.H.S=1.4÷[4.2.3]

R.H.S=1/6

P(n) is true for n=1, hence proved!!

Assume P(k) is true

1÷1.2.3+1÷2.3.4+1÷3.4.5+....+1÷k(k+1)(k+2)=k(k+3)÷4(k+1)(k+2)

L.H.S

=1/1.2.3+1/2.3.4+.........1/[(k+1)((k+1)+1)((k+1)+2)]

=1/1.2.3+1/2.3.4+.....+1/k(k+1)(k+2)=k(k+3)/4(k+1)(k+2)

=k(k+3)/4(k+1)(k+2)+1/(k+1)(k+2)(k+3)

=1/(k+1)(k+2)*[k(k+3)/4+1/(k+3)]

=1/(k+1)*[k(k^2+3^2+2(3)(k))+4/(k+2)(k+3)]

=1/(k+1)(k+2)*[k(k^2+3^2+6k)+4/4(k+3)]

=1/(k+1)(k+2)*[k^3+9k+6k^2+4/4(k+3)]

=(k+1)(k^2+5k+4)/4(k+1)(k+2)(k+3)

=(k+1)[k(k+4)+1(k+4)]

=(k+1)(k+4)/4(k+2)(k+3)

R.H.S

=(K+1)[(K+1)+3]÷4[(K+1)+1]*[(K+1)+2]

=(k+1)(k+4)/4(k+2)(k+3)

L.H.S=R.H.S

Therefore, P(k+1) is true for when P(k) is true.

By mathematical induction, P(n) is true when n is natural number.