anonymous
  • anonymous
Use mathematical induction to prove that: 1/1*2+1/2*3+1/3*4+...+1/n*(n+1)=n/n+1 Have no idea where to start!
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
well... one would think at the base case :P
anonymous
  • anonymous
so its obviously having to prove it for n>1
anonymous
  • anonymous
n>=1

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anonymous
  • anonymous
so check n=1 , its true , thats the first step done
anonymous
  • anonymous
then assume it is true for n=k , ie that S(k) { the sum of k terms } = k / (k+1)
anonymous
  • anonymous
now we need to prove it is true for n=k+1 , ie that S (k+1) = (k+1)/ ( k+2) where we got the RHS , by substituting n=k+1 in the summation
anonymous
  • anonymous
now there is a result that S(k+1) = S(k) + T(k+1) that is , in words, "the sum of k+1 terms is equal to sum of the first k terms , plus the (k+1)th term "
anonymous
  • anonymous
like saying, sum 1 + 2 + 3 + 4 , well thats equal to summing the first 3 ( to get 6 ) and then adding on 4 ( the (k+1)th term )
anonymous
  • anonymous
now , back to the question , S(k+1) = k / (k+1) + 1/ (k+1)(k+2)
anonymous
  • anonymous
S(k+1) = [ k/(k+1) ] + 1/ [ (k+1)(k+2)] where the first fraction came from our assumption , that is what S(k) was assumed to be ! and the second term comes from subing n=k+1 into the general term on the LHS of the summation
anonymous
  • anonymous
now if you get S(k+1) over a single denominator and simplify , you will get S(k+1) = [ (k+1)/(k+2)] , which is what we set out to prove from step 3. Hence it is proved by induction the end
anonymous
  • anonymous
oh, my! thank you! I would never have gotten this. Thank you for explaining it step by step. SOOOOOO helpful!

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