Question

1/1×2+1/2×3+1/3×4+1/4×5+1/5×6+1/6×7+1/7×8+1/8×9+1/9×10=?​

Answer 1

=(1-1/2)+(1/2-1/3)+(1/3-1/4)+.........+(1/9-1/10)

=1-1/10=9/10

Answer 2

First of all, let me find whether there is any method to find such terms up to a particular term or not.

Here, the denominator seems as product of position number and 1 added to it. Thus we get a clear picture that the denominator is in the form n(n + 1). So, let me find the algebraic expression for such sums.

For this, I'm finding it by something like mathematical induction.

When we consider n = 1, it's 1/2, where numerator is n = 1 and denominator is 1 greater than numerator.

When we consider n = 2, it's 1/2 + 1/6 = 2/3, where also numerator is n = 2 and denominator is n + 1 = 3.

When we consider n = 3, it's 1/2 + 1/6 + 1/12 = 3/4, where also numerator is n = 3 and denominator is n + 1 = 4.

So let me consider n = k and assume that the numerator of 1/2 + 1/6 + 1/12 + ... + 1/k(k + 1) is n = k and the denominator is n + 1 = k + 1.

Then let n = k + 1, and the calculation here is given below:

$\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+...+\frac{1}{(k+1)(k+1+1)} \\ \\ \frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+...+\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)} \\ \\ \frac{k}{k+1}+\frac{1}{(k+1)(k+2)} \\ \\ \frac{k(k+1)(k+2)+(k+1)}{(k+1)(k+1)(k+2)}\ \ \ \ \ \ \ \ \ \ [\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}] \\ \\ \frac{(k+1)(k(k+2)+1)}{(k+1)^2(k+2)} \\ \\ \frac{(k+1)(k^2+2k+1)}{(k+1)^2(k+2)} \\ \\ \frac{(k+1)(k+1)^2}{(k+1)^2(k+2)} \\ \\ \frac{k+1}{k+2}$

Here also seems that the numerator is n = k + 1 and he denominator is n + 1 = k + 2.

Thus we can conclude the proof with the concept that,

$\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+...+\frac{1}{n(n+1)}=\frac{n}{n+1}$

So let's answer the question with this identity.

The question is,

$\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+\frac{1}{4 \times 5}+\frac{1}{5 \times 6}+\frac{1}{6 \times 7}+\frac{1}{7 \times 8}+\frac{1}{8 \times 9}+\frac{1}{9 \times 10}$

Here let that n = 9. Such that the answer is,

$\bold{\frac{9}{10}}$

So the answer is 9/10.

So, remember the identity that,

$\displaystyle \frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+...+\frac{1}{n(n+1)}=\frac{n}{n+1}$

which can also be written as,

$\displaystyle\sum_{k=1}^{n}\frac{1}{k(k+1)} \ = \ \frac{n}{n+1}$