FIND THE SUM OF SPECIAL SERIES

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)² -  (20/2)²

= 81² - 10²

= (81+10)(81-10)

= 91(71)

= 6461