1> 1\2.5+1\5.8+......+1\(3n-1)(3n+2)=n\(6n+4)

Asked by visakha | 30th Nov, 2013, 06:22: PM

Expert Answer:

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

Let us 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.

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)

Now, we will 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.

Thus, by the principal of mathematical induction we can say that S(n) is true for all natural numbers n.

Answered by  | 1st Dec, 2013, 01:15: AM

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.