Mathematics SEQUENCES AND SERIES - Sum to n Terms of Special Series
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### Sum to n Terms of Special Series

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.

### sum of first n natural numbers

(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))]

### sum of squares of the first n natural numbers

(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)

### sum of cubes of the first n natural numbers

(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)

Q 3079723616

Find the sum to n terms of the series: 5 + 11 + 19 + 29 + 41…

Solution:

Let us write

S_n = 5 + 11 + 19 + 29 + .........+a_(n-1) +a_n

S_n = \ \ \ \ \ \ \ \ \ \ 5+11+19+29............+a_(n-1) +a_n

On subtraction, we get

0 = 5 + [6 + 8 + 10 + 12 + ...(n – 1) "terms"] – a_n

or a_n = 5+ {(n-1) [12+(n-1) xx 2])/2

= 5+(n-1) (n+4) = n^2+3n+1

Hence S_n = underset(k =1) overset(n)Sigma a_k = underset(k =1) overset(n)Sigma (k^2+3k+1) = underset(k =1) overset(n)Sigma k^2+3underset(k =1) overset(n)Sigma k +n

 = { n(n+1)(2n+1)}/6+{3n(n+1)}/2 +n = (n(n+2)(n+4))/3
Q 3029823711

Find the sum to n terms of the series whose nth term is n (n+3).

Solution:

Given that a_n = n (n + 3) = n^2 + 3n
Thus, the sum to n terms is given by

S_n = underset(k=1) overset(n) Sigma a_k = underset(k=1) overset(n) Sigma k^2 + 3 underset(k=1) overset(n) Sigma k

 = (n(n+1)(2n+1))/6+(3n(n+1))/2

 = (n(n+1)(n+5))/3