FIND THE SUM OF SPECIAL SERIES

Special series comes under one of the following category.

(i) Sum of natural numbers

(ii)  Sum of squares

(iii) Sum of cubes

(iv) Sum of odd numbers

Find the sum of following special series.

Problem 1 :

1 + 2 + 3 + … + 20

Solution :

Sum of natural numbers :

1 + 2 + 3 + … + n = n(n + 1)/2

n = 20

1 + 2 + 3 + … + 20 = 20(20 + 1)/2

= 10(21)

= 210

Problem 2 :

12 + 22 + 32 + … + 102

Solution :

Sum of squares natural numbers :

12 + 22 + 32 + … + n2 = n(n + 1)(2n + 1)/6

n = 10

= [10(10 + 1)(2(10) + 1)]/6

= 10(11)(20 + 1)/6

= 10(11)(21)/6

= 2310/6

= 385

Problem 3 :

2 + 8 + 18 + … + (2 × 152)

Solution :

Given, 2 + 8 + 18 + … + 450

2[1 + 4 + 9 + … + 225]

2[12 + 22 + 32 + …+ 152]

Sum of squares natural numbers :

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

n = 15

12 + 22 + 32 + … + 152 = 2 × 15(15 + 1)(2(15) + 1)/6

= 30(16)(31)/6

= 2480

Problem 4 :

2 + 4 + 6 + … + 100

Solution :

Sum of natural numbers :

= 2 + 4 + 6 + … + 100

Factoring 2 from all, we get

= 2(1 + 2 + 3 + … + 50)

n = 50

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

= 50(51)

= 2550

Problem 5 :

1 + 3 + 5 + … + 25

Solution :

Since we dont know the total number of terms, we can use the formula

= [(l + 1)/2]2

= [(25 + 1)/2]2

= 132

= 169

Problem 6 :

1 + 8 + 27 + … + 1000

Solution :

Sum of cubes natural numbers :

1 + 8 + 27 + … + 1000 = 13 + 23 + 33 + … + 103

13 + 23 + 33 + … + n3 = [n(n + 1)/2]2

n = 10

= [10(10 + 1)/2]2

= [5(11)]2

= (55)2

= 3025

Problem 7 :

112 + 122 + … + 242

Solution :

112 + 122 + … + 242

= (12 + 22 + … + 242) – (12 + 22 + … + 102)

n = 24 & n = 10

Sum of squares = n(n + 1)(2n + 1)/6

= [24(24 + 1)(2(24) + 1)/6] – [10(10 + 1)(2(10) + 1)/6]

= [24(25)(49)/6] – [10(11)(21)/6]

= 29400/6 – 2310/6

= 4900 – 385

= 4515

Problem 8 :

73 + 83 + … + 153

Solution :

73 + 83 + … + 153

= (13 + 23 + … + 153) – (13 + 23 + … + 63)

n = 15 & n = 6

Sum of cubes = [n(n + 1)/2]2

= [15(15 + 1)/2]2 – [6(6 + 1)/2]2

= [15(16)/2]2 – [6(7)/2]2

= [15(8)]2 – [3(7)]2

= [120]2 – [21]2

= (120 + 21) (120 – 21)

= 141(99)

= 13959

Problem 9 :

21 + 23 + … + 61

Solution :

21 + 23 + … + 61

= (1 + 3 + 5 + … + 61) – (1+ 3 + 5 … + 19)

l = 61 & l = 19

= [(61 + 1)/2]2 - [(19 + 1)/2]2

= (62/2)2 - (20/2)2

= 312 - 102

= (31 + 10) (31 - 10)

= 41 (21)

= 861

Problem 10 :

Calculate 21 + 23 + 25 … + 161

Solution :

21 + 23 + 25 … + 161

= (1 + 3 + 5 … + 161) – (1 + 3 + 5 + … + 19)

n = 161 & n = 19

= [(161 + 1)/2]2 -  [(19 + 1)/2]2

= (162/2)-  (20/2)2

= 81- 102

= (81+10)(81-10)

= 91(71)

= 6461

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