Question #80860

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Question #80860

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Answer 1 · 2016-09-03T20:43:16

Notice that the given be written as follows

$(1/1-1/2)+(1/2-1/3)+(1/3-1/4)+...+(1/n-1/(n+1))$

regroup and we have

$(1/1)+(-1/2+1/2)+(-1/3+1/3)+(-1/n+1/n)-1/(n+1)$

The sums inside the parentheses are zero hence we left with

$(1/1-1/((n+1)))=(n+1-1)/(n+1)=n/(n+1)$

Answer 2 · 2016-09-03T21:10:09

See below.

Explanation:

For $n=1$ is true for

$1/(1 xx 2) = 1/2$

Supposing that is true for $n$

$sum_(k=1)^n1/(k(k+1)) = n/(n+1)$

then it must be true for $n+1$

$sum_(k=1)^(n+1)1/(k(k+1)) = sum_(k=1)^n1/(k(k+1)) + 1/((n+1)(n+2))$
$= n/(n+1)+ 1/((n+1)(n+2)) = (n(n+2))/((n+1)(n+2))+ 1/((n+1)(n+2))$
$=(n+1)^2/((n+1)(n+2)) = (n+1)/(n+2)$

then by mathematical induction, the affirmation is true.

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Question #80860

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