Question

1/2.5+1/5.8+1/8.11+...+1/(3n-1)(3n+2)=n/6n+4

prove the following by using the principal of mathematical induction for all n€N
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Answer 1

here is your answer OK

In this step follow......

Let S(n) : 1/2.5+1/5.8+1/8.11 +.... + 1/(3n-1)(3n+2) = n/(6n+4)

Step-1:

Prove S(n) for n=1.

LHS =1/2.5 =1/10; RHS = 1/6+4=1/10

So it is true for n=1.

Step-2:

Assume that S(n) is true for some n=k.

1/2.5+1/5.8+1/8.11 +.... + 1/(3k-1)(3k+2) = k/(6k+4)

Step-3:

Prove S(n) for n=k+1

LHS = 1/2.5+1/5.8+1/8.11 +.... + 1/(3(k+1)-1)(3(k+1)+2)

=k/(6k+4) + 1/(3k+2)(3k+5)

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

=[k(3k+5)+2]/[2((3k+2)(3k+5)]

=[3k^2+5k+2]/[2((3k+2)(3k+5)]

=[(3k+2)(k+1)]/[2((3k+2)(3k+5)]

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

=(k+1)/(6k+10)

= (k+1)/[6(k+1)+4]

=RHS

So it is true for n=k+1.

So, by the principal of mathematical induction we can say that it is true for all natural numbers n.

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