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)2 - (20/2)2
= 812 - 102
= (81+10)(81-10)
= 91(71)
= 6461
Sep 22, 23 08:41 AM
Sep 22, 23 06:13 AM
Sep 22, 23 06:09 AM