1 +1/1+2 +1/1+2+3 +.....+1/1+2+3+...n =2n/n+1.prove by principle of mathematical induction.

Asked by Akash Sundi | 15th Mar, 2012, 03:06: PM

Expert Answer:

P(n) = 1  + 1/(1+2) + ...... + 1/(1+2+3+ ... +n)
So P(1) = 1 , Put n =1 in 2n/(n+1) we get 1
thus P(1) is true , Now let P(k) is true
So P(k) =  1  + 1/(1+2) + ...... + 1/(1+2+3+ ... +k)
To Prove is P(k+1) is true that is P(k+1) = 2(k+1)/(k+2)
Now P(k+1) =  1  + 1/(1+2) + ...... + 1/(1+2+3+ ... +k) + 1/(1+2+3+ ... k+k+1)
= 2k/(k+1) + 1/{(k+2)(k+1)/2}    [ sum of n terms = n(n+1)/2]
= 2k/(k+1) + 2/(k+2)(k+1)
= 2{k2 + 2k + 1}/(k+1)(k+2) = 2(k+1)/(k+2) = R.H.S.

Answered by  | 15th Mar, 2012, 05:38: PM

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