# SUM TO n TERMS OF SPECIAL SERIES

## Sum of First n Natural Numbers

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

Example 1 :

Find the sum :

1 + 2 + 3 + ............ + 60

Solution :

Sum of first n natural numbers :

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

Substitute n = 60.

1 + 2 + 3 + ............ + 60 = 60(60 + 1)/2

= 30(61)

= 1830

Example 2 :

Find the sum :

3 + 6 + 9 + ............ + 96

Solution :

= 3 + 6 + 9 + ............ + 96

= 3(1 + 2 + 3 + ............ + 32)

= 3[32(32 + 1)/2]

= 3(16)(33)

= 1584

Example 3 :

Find the sum :

51 + 52 + 53 + ............ + 92

Solution :

51+ 52 + 53 + ............ + 92 :

= (1 + 2 + 3 + ............ + 92) - (1 + 2 + 3 + ............ + 50)

= [92(92 + 1)/2] - [50(50 + 1)/2]

= 46(93) - 25(51)

= 4278 - 1275

= 3003

## Sum of Squares of first n Natural Numbers

12 + 22 + 32 + ............ + n2 = [n(n + 1)(2n + 1)]/6

Example 4 :

Find the sum :

1 + 4 + 9 + 16 + ............ + 225

Solution :

1 + 4 + 9 + 16 + ............ + 225 :

= 12 + 22 + 32 + 42 + ............ + 152

Sum of squares of first n natural numbers :

12 + 22 + 32 + 42 + ............ + n2 = [n(n + 1)(2n + 1)]/6

Substitute n = 15

12 + 22 + 32 + 42 + ............ + 152 :

= [15(15 + 1)(2(15) + 1)]/6

= 15(16)(31)/6

= 5(8)(31)

= 1240

Example 5 :

Find the sum :

62 + 72 + 82............ +212

Solution :

62 + 72 + 82 + ............ + 212 :

= (12 + 22 + ............ + 212) - (12 + 22 + ............ + 52)

= [21(21 + 1)(2(21) + 1)/6] - [5(5 + 1)(2(5) + 1)/6]

= [21(22)(43)/6] - [5(6)(11)/6]

= [7(11)(43)] - [5(1)(11)]

= 3311 - 55

= 3256

## Sum of Cubes of First n Natural Numbers

13 + 23 + 33 + ............ + n3 = [n(n + 1)/2]2

Example 6 :

103 + 113 + 123 + ............ + 203

Solution :

103 + 113 + 123 + ............ + 203 :

= (13 + 23 + 33 +.........+ 203) - (13 + 23 + 33 +.........+ 93)

Sum of cubes of first n natural numbers :

13 + 23 + 33 +.........+ n3 = [n(n + 1)/2]2

Then,

= [20(20 + 1)/2]2 - [9(9 + 1)/2]2

= 2102 - 452

= (210 + 45)(210 - 45)

= 255(165)

= 42075

## Sum of first n Odd Numbers

When number of terms n is given :

1 + 3 + 5 + ............ to n terms = n2

When the last term l is given :

1 + 3 + 5 + ............ + = [(l + 1)/2]2

Example 7 :

1 + 3 + 5 + ............ to 25 terms

Solution :

Sum of first n odd numbers :

1 + 3 + 5 + ............ + n = n2

Substitute n = 25.

= 252

= 625

Example 8 :

1 + 3 + 5 + ............ + 71

Solution :

Sum of first n odd numbers :

1 + 3 + 5 + ............ + = [(l + 1)/2]2

Substitute = 71.

= [(71 + 1)/2]2

= [72/2]2

= 362

= 1296

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