# Summation of series(method of differences),When we use ∑f(r)-f(r+1),how can we know it is f(1)-f(n+1), but not f(n)-f(1+1)? Also, how can we know it is f(n+1)-f(1) but not f(1+1)-f(n) when we use ∑f(r+1)-f(r)?

##### 2 Answers
Jul 18, 2017

${\sum}_{r = 1}^{n} f \left(r\right) - f \left(r + 1\right) = f \left(1\right) - f \left(n + 1\right)$

${\sum}_{r = 1}^{n} f \left(r + 1\right) - f \left(r\right) = f \left(n + 1\right) - f \left(1\right)$

#### Explanation:

Write out the terms and see what happens:

Consider:

${S}_{1} = {\sum}_{r = 1}^{n} f \left(r\right) - f \left(r + 1\right)$

$\text{ } = \left\{f \left(1\right) - f \left(2\right)\right\} +$
$\text{ } \left\{f \left(2\right) - f \left(3\right)\right\} +$
$\text{ } \left\{f \left(3\right) - f \left(4\right)\right\} +$
$\text{ } \left\{f \left(4\right) - f \left(5\right)\right\} + \ldots +$

$\text{ } \left\{f \left(n - 1\right) - f \left(n\right)\right\} +$
$\text{ } \left\{f \left(n\right) - f \left(n + 1\right)\right\}$

$\text{ } = \left\{f \left(1\right) - \cancel{\textcolor{p u r p \le}{f} \left(2\right)}\right\} +$
$\text{ } \left\{\cancel{\textcolor{p u r p \le}{f} \left(2\right)} - \cancel{\textcolor{red}{f} \left(3\right)}\right\} +$
$\text{ } \left\{\cancel{\textcolor{red}{f} \left(3\right)} - \cancel{\textcolor{g r e e n}{f} \left(4\right)}\right\} +$
$\text{ } \left\{\cancel{\textcolor{g r e e n}{f} \left(4\right)} - \cancel{f \left(5\right)}\right\} + \ldots +$

$\text{ } \left\{\cancel{f \left(n - 1\right)} - \cancel{\textcolor{b l u e}{f} \left(n\right)}\right\} +$
$\text{ } \left\{\cancel{\textcolor{b l u e}{f} \left(n\right)} - f \left(n + 1\right)\right\}$

And as shown almost all terms vanish, leaving:

${S}_{1} = f \left(1\right) - f \left(n + 1\right)$

Whereas with:

${S}_{2} = {\sum}_{r = 1}^{n} f \left(r + 1\right) - f \left(r\right)$

$\text{ } = \left\{f \left(2\right) - f \left(1\right)\right\} +$
$\text{ } \left\{f \left(3\right) - f \left(2\right)\right\} +$
$\text{ } \left\{f \left(4\right) - f \left(3\right)\right\} +$
$\text{ } \left\{f \left(5\right) - f \left(4\right)\right\} + \ldots +$

$\text{ } \left\{f \left(n\right) - f \left(n - 1\right)\right\} +$
$\text{ } \left\{f \left(n + 1\right) - f \left(n\right)\right\}$

$\text{ } = \left\{\cancel{\textcolor{p u r p \le}{f} \left(2\right)} - f \left(1\right)\right\} +$
$\text{ } \left\{\cancel{\textcolor{red}{f} \left(3\right)} - \cancel{\textcolor{p u r p \le}{f} \left(2\right)}\right\} +$
$\text{ } \left\{\cancel{\textcolor{g r e e n}{f} \left(4\right)} - \cancel{\textcolor{red}{f} \left(3\right)}\right) +$
$\text{ } \left\{\cancel{f \left(5\right)} - \cancel{\textcolor{g r e e n}{f} \left(4\right)}\right\} + \ldots +$

$\text{ } \left\{\cancel{\textcolor{b l u e}{f} \left(n\right)} - \cancel{f \left(n - 1\right)}\right\} +$
$\text{ } \left\{f \left(n + 1\right) - \cancel{\textcolor{b l u e}{f} \left(n\right)}\right\}$

Similarly, almost all terms vanish, leaving:

${S}_{2} = - f \left(1\right) + f \left(n + 1\right)$
$\text{ } = f \left(n + 1\right) - f \left(1\right)$

Jul 18, 2017

Refer to the Explanation.

#### Explanation:

Let us say, $S = {\sum}_{r = 1}^{n} \left\{f \left(r\right) - f \left(r + 1\right)\right\} .$ To find $S ,$ we substitute,

$r = 1 , 2 , 3 , \ldots , n - 1 , n ,$ successively in $\left\{f \left(r\right) - f \left(r + 1\right)\right\} ,$ and get,

$S = \left\{f \left(1\right) - \cancel{f} \left(2\right)\right\} + \left\{\cancel{f} \left(2\right) - \cancel{f} \left(3\right)\right\} + \left\{\cancel{f} \left(3\right) - \cancel{f} \left(4\right)\right\}$

$+ \ldots + \left\{\cancel{f} \left(n - 1\right) - \cancel{f} \left(n\right)\right\} + \left\{\cancel{f} \left(n\right) - f \left(n + 1\right)\right\} ,$

$\Rightarrow S = {\sum}_{r = 1}^{n} \left\{f \left(r\right) - f \left(r + 1\right)\right\} = f \left(1\right) - f \left(n + 1\right) .$

This proves the ${1}^{s t}$ Assertion.

The Proof of the ${2}^{n d}$ Assertion can be obtained similarly.

Otherwise, by what we have proved above,

${\sum}_{r = 1}^{n} \left\{f \left(r\right) - f \left(r + 1\right)\right\} = f \left(1\right) - f \left(n + 1\right) .$

Multiplying this by $- 1 ,$ we have,

$- {\sum}_{r = 1}^{n} \left\{f \left(r\right) - f \left(r + 1\right)\right\} = - \left\{f \left(1\right) - f \left(n + 1\right)\right\} , i . e . ,$

$\therefore {\sum}_{r = 1}^{n} \left\{f \left(r + 1\right) - f \left(r\right)\right\} = f \left(n + 1\right) - f \left(1\right) .$

Enjoy Maths.!