Question

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)`?

Answer 1

`sum_(r=1)^n f(r)-f(r+1) = f(1)-f(n+1)`

`sum_(r=1)^n f(r+1)-f(r) = f(n+1) -f(1)`

Explanation:

Write out the terms and see what happens:

Consider:

`S_1 = sum_(r=1)^n f(r)-f(r+1)`

`" " = {f(1)-f(2)} +`
`" " {f(2)-f(3)} +`
`" " {f(3)-f(4)} +`
`" " {f(4)-f(5)} + ... +`

`" " {f(n-1)-f(n)} +`
`" " {f(n)-f(n+1)}`

`" " = {f(1)-cancel(color(purple)f(2))} +`
`" " {cancel(color(purple)f(2))-cancel(color(red)f(3))} +`
`" " {cancel(color(red)f(3))-cancel(color(green)f(4))} +`
`" " {cancel(color(green)f(4))-cancel(f(5))} + ... +`

`" " {cancel(f(n-1))-cancel(color(blue)f(n))} +`
`" " {cancel(color(blue)f(n))-f(n+1)}`

And as shown almost all terms vanish, leaving:

`S_1 = f(1)-f(n+1)`

Whereas with:

`S_2 = sum_(r=1)^n f(r+1)-f(r)`

`" " = {f(2)-f(1)} +`
`" " {f(3)-f(2)} +`
`" " {f(4)-f(3)} +`
`" " {f(5)-f(4)} + ... +`

`" " {f(n)-f(n-1)} +`
`" " {f(n+1)-f(n)}`

`" " = {cancel(color(purple)f(2))-f(1)} +`
`" " {cancel(color(red)f(3))-cancel(color(purple)f(2))} +`
`" " {cancel(color(green)f(4))-cancel(color(red)f(3)}) +`
`" " {cancel(f(5))-cancel(color(green)f(4))} + ... +`

`" " {cancel(color(blue)f(n))-cancel(f(n-1))} +`
`" " {f(n+1)-cancel(color(blue)f(n))}`

Similarly, almost all terms vanish, leaving:

`S_2 = -f(1) + f(n+1)`
`" " = f(n+1) -f(1)`

Answer 2

Refer to the Explanation.

Explanation:

Let us say, `S=sum_(r=1)^n{f(r)-f(r+1)}.` To find `S,` we substitute,

`r=1,2,3,...,n-1,n,` successively in `{f(r)-f(r+1)},` and get,

`S={f(1)-cancelf(2)}+{cancelf(2)-cancelf(3)}+{cancelf(3)-cancelf(4)}`

`+...+{cancelf(n-1)-cancelf(n)}+{cancelf(n)-f(n+1)},`

`rArr S=sum_(r=1)^n{f(r)-f(r+1)}=f(1)-f(n+1).`

This proves the `1^(st)` Assertion.

The Proof of the `2^(nd)` Assertion can be obtained similarly.

Otherwise, by what we have proved above,

`sum_(r=1)^n{f(r)-f(r+1)}=f(1)-f(n+1).`

Multiplying this by `-1,` we have,

`-sum_(r=1)^n{f(r)-f(r+1)}=-{f(1)-f(n+1)}, i.e.,`

`:. sum_(r=1)^n{f(r+1)-f(r)}=f(n+1)-f(1).`

Enjoy Maths.!