We shall now find the sum of first `n` terms of some special series, namely;

`(i) \ \ color(blue)(1 + 2 + 3 +… + n) ` (sum of first `n` natural numbers)

`(ii) \ \ color(blue)(1^2 + 2^2 + 3^2 +… + n^2)` (sum of squares of the first `n` natural numbers)

`(iii) \ \ color(blue)(1^3 + 2^3 + 3^3 +… + n^3) ` (sum of cubes of the first `n` natural numbers).

Let us take them one by one.

(i) `color(red)( S_n = 1+2+3 + ...............+ n)`

Clearly, it is an Arithmetic Progression whose `color(blue)("first term = 1")`, `color(blue)("last term = n")` and `color(blue)("number of terms = n.")`

Therefore,`color(red)(S_n = (n(n+1))/2)` , [Using the formula `color(blue)(S = n/2(a + l))`]

(ii) Here `color(red)(S_n = 1^2+2^2+3^2 +..........+ n^2)`

We consider the identity `color(blue)(k^3 – (k – 1)^3 = 3k^2 – 3k + 1)`

Putting `color(blue)(k = 1, 2......…) `successively, we obtain

`1^3 – 0^3 = 3 (1)^2 – 3 (1) + 1`

`2^3 – 1^3 = 3 (2)^2 – 3 (2) + 1`

`3^3 – 2^3 = 3(3)^2 – 3 (3) + 1`
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`n^3 – (n – 1)^3 = 3 (n)^2 – 3 (n) + 1`

`color(green)("Adding both sides")`, we get

`color(purple)(n^3 – 0^3 = 3 (1^2 + 2^2 + 3^2 + ... + n^2) – 3 (1 + 2 + 3 + ... + n) + n)`

`color(blue)(n^3 = 3 underset(k= 1) overset(n)Sigma k^2-3 underset(k=1) overset(n) Sigma k+n)`

By (i), we know that `underset(k= 1) overset(n) k =1+2+3 + ...............+ n =(n(n+1))/2 `

Here `color(red)(underset(k=1) overset(n)Sigma k^2 )= 1/3 [ n^3 + (3n(n+1))/2-n] = 1/6 (2n^3+3n^2+n)`

` \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ color(red)(S_n= (n(n+1)(2n+1))/6)`

(iii) Here `color(red)(S_n = 1^3+2^3+..........+ n^3)`

We consider the identity, `color(blue)((k + 1)^4 – k^4 = 4k^3 + 6k^2 + 4k + 1)`

Putting `color(blue)(k = 1, 2, 3… n)`, we get

`2^4 – 1^4 = 4(1)^3 + 6(1)^2 + 4(1) + 1`

`3^4 – 2^4 = 4(2)^3 + 6(2)^2 + 4(2) + 1`

`4^4 – 3^4 = 4(3)^3 + 6(3)^2 + 4(3) + 1`
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`(n – 1)^4 – (n – 2)^4 = 4(n – 2)^3 + 6(n – 2)^2 + 4(n – 2) + 1`

`n^4 – (n – 1)^4 = 4(n – 1)3 + 6(n – 1)2 + 4(n – 1) + 1`

`(n + 1)^4 – n^4 = 4n^3 + 6n^2 + 4n + 1`

`color(green)("Adding both sides")`, we get

`color(purple)((n + 1)^4 – 1^4 = 4(1^3 + 2^3 + 3^3 +...+ n^3) + 6(1^2 + 2^2 + 3^2 + ...+ n^2) + 4(1 + 2 + 3 +...+ n) + n)`


` \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ color(blue)(= 4 underset(k=1) overset(n)Sigma k^3+6underset(k=1) overset(n)Sigmak^2+4 underset(k=1) overset(n)Sigma k+n)` .......(1)


From parts `(i)` and `(ii),` `color(blue)("we know that")`

`color(red)(underset(k=1) overset(n)Sigma k = (n(n+1))/2)` and `color(red)(underset(k=1) overset(n) Sigma k^2 = (n(n+1) (2n+1))/6)`

Putting these values in equation `(1)`, we obtain

`color(red)(4 underset(k=1) overset(n) Sigma k^3) = n^4+4n^3+6n^2+4n - (6n(n+1)(2n+1))/6 - (4n (n+1))/2-n`

or `color(red)(4S_n )= n^4+4n^3+6n^2+4n-n(2n^2+3n+1) - 2n (n+1) -n`

`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \= n^4+2n^3+n^2`

`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = n^2 (n+1)^2`

Hence ` color(red)(S_n = (n^2(n+1)^2)/4 = ([ n(n+1)]^2)/4)`