# Evaluate the sum 1/2+1/6+...+1/k(k+1)

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To find the sum 1/2 +1/6+..1/k(k+1).

We know that 1/k(k+1) = 1/k-1/(k+1).

Therefore the term by term we can write as below:

1/2 = (1//1-1/2).

1/6 = (1/2-1/3).

1/12 = (1/3-1/4).

1/(20 = (1/4 -1/5).

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1/k9k+1) = 1/k-1/(k+1)

Adding the k terms we get:

1/2 +1/6+1/12+1/20 +....1/(k+1) = 1/2 - 1/(k+1), as all other terms on the right cancel while adding.

Therefore the sum (1/2+1/6+1/12+1/20 ...1/k(k+1) = **1/2-1/(k+1)** = (2k+2-2)/2(k+1)= 2k/2(k+1) =** k/(k+1)**

The denominator of each fraction is the product of 2 consecutive numbers.

1/1*2 + 1/2*3 + ....+ 1/k(k+1)

We'll write this type of ratio 1/n(n+1) as the result of addition or subtraction of 2 elementary fractions:

We'll write the ratio:

1/n(n+1) = A/n + B/(n+1) (1)

We'll multiply the ratio A/n by (n+1) and we'll multiply the ratio B/(n+1) by n.

1/n(n+1) = [A(n+1) + Bn]/n(n+1)

We'll write the numerators, only.

1 = A(n+1) + Bn

We'll remove the brackets:

1 = An + A + Bn

We'll factorize by n to the right side:

1 = n(A+B) + A

We'll put equal the correspondent coefficientsfrom both sides.

A+B = 0

A = 1

1 + B = 0

B = -1

We'll substitute A and B into the expression (1):

**1/n(n+1) = 1/n - 1/(n+1) (2)**

According to (2), we'll give values to n:

for n = 1 => 1/1*2 = 1/1 - 1/2

for n = 2 => 1/2*3 = 1/2 - 1/3

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for n = k-1 => 1/k(k-1) = 1/(k-1) - 1/k

for n = k => 1/k(k+1) = 1/k - 1/(k+1)

If we'll add the ratios from the left side, we'll get:

1/1*2 + 1/2*3 + ... + 1/k(k-1) + 1/k(k+1) = 1/1 - 1/2 + 1/2 - 1/3 + ..... + 1/(k-1) - 1/k + 1/k - 1/(k+1)

We'll eliminate like terms from the right side:

1/1*2 + 1/2*3 + ... + 1/k(k-1) + 1/k(k+1) = 1 - 1/(k+1)

S = (k + 1 - 1)/(k+1)

We'll eliminate like terms and we'll get the final result of the sum:

**S = k /(k+1)**