I just started this subject, and I like to know how to solve excersise like this. I know the properties.

Question

I just started this subject, and I like to know how to solve excersise like this. I know the properties.

$ \sum_{k=1}^{100} (\frac{1}{k}-\frac{1}{k+1}) $

Answer 1

It's a trick. But it's a very common and well known trick that everyone is expected to know.

$\sum_{k=1}^{100} (\frac{1}{k}-\frac{1}{k+1}) $

$= \sum_{k=1}^{100}\frac{1}{k} - \sum_{k=1}^{100}\frac{1}{k+1}$

$ = (\sum_{k=0}^{99}\frac{1}{k + 1}) - (\sum_{k=1}^{100}\frac{1}{k+ 1}) $

$= (\frac{1}{k+1; k = 0} + \sum_{i=1}^{99}\frac{1}{k+1}) - (\sum_{i=1}^{99}\frac{1}{k+ 1} + \frac{1}{k + 1; k = 100})$

$= \frac{1}{0 + 1} + (\sum_{i=1}^{99}\frac{1}{k+1} -\sum_{i=1}^{99}\frac{1}{k+ 1}) - \frac{1}{100 + 1} $

$=1 - \frac{1}{101} = \frac{100}{101}$

In other words, make it a difference (or sum) of two sums, reindex to make the terms compatible, recombine and simplify.

It's exactly the same concept if $(x - 1)(x^n + x^{n-1} + .... + x + 1) = x^n - 1$ for exactly the same reason-- you get a sequence of cancelling differences which each are equal to zero, so only the "extreme" terms remain.

In general $\sum_{i=1}^{n}(a_{i+1} - a_i) = a_{n+1} - a_1$. (Think about it. $(a_2 - a_1) + (a_3 - a_2) + (a_4 - a_3) +...... $)

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It's used all the time. For example: $(x - 1)(x^n + x^{n-1} + .... + x + 1) = (x^{n+1} - x^n) + (x^n - x^{n-1}) + .... + (x^2 - x) + (x - 1) = x^{n+1} - (x^n - x^n) -(x^{n-1} - x^{n-1}) + .... - (x - x) - 1 = x^{n-1} - 1$

so $(x^n + x^{n-1} + .... + x + 1) = \frac{x^{n+1} - 1}{x-1}$ (for x $\ne$ 1). [If x = 2 that is a very well known result you have probably known since the 4th grade!]

Answer 2

Hint

Write your sum as $$\sum_{i=1}^{100}a_i-a_{i+1}=(a_i-a_{i+1})+(a_{i+1}-a_{i+2})+...$$

After finding the pattern, plug in the appropiate value for $a_i$.