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Evaluate the series: (a) sum_{k=1}^{20) k (b) sum_{j = 1}^{60} 5 j^2 (c) sum_{i = 1}^{11} 2ln3-5i

  • Last Updated : 07 Jan, 2022

The number system is a concept of representing numbers on a defined number line using digits (0 – 9) and a primitive set of rules over the numbers. Number system can be of many types like,

  1. Decimal Number System (Base – 10 , using digits (0 – 9))
  2. Binary Number System (Base – 2, using digits 0 and 1)
  3. Octal Number System (Base – 8, using digits (0 – 7)
  4. Hexa-Decimal Number System (Base-16, using digits (0 – 9) and alphabets (A – F)

Sequences and Series 

A number series is a sequence/series of numbers. In the questions regarding number series, there can be a pattern or rules applied to the numbers to obtain a given series. So, a number series can be the product of a sequence of numbers or a  sum of the sequence of numbers. Examples:

  1. Sum of first 10 natural numbers
  2. Sum of first 10 even numbers
  3. Sum of first 10 odd numbers
  4. Product of first 10 natural numbers.

These series can be solved by using various techniques such as Arithmetic Progression (AP), Geometric Progression (GP), Arithmetic-Geometric Progression (AGP), least squares estimation, regression algorithms, etc.., But most fundamental techniques to solve a sequence are Arithmetic Progression (AP) and Arithmetic-Geometric Progression (AGP). Some Arithmetic Progression and Geometric Progression formulas can be seen below.

Arithmetic Progression Formulae

ar = a1 + (r – 1)d

Sn = n/2(2a + (n – 1)d)

n – number of elements in a series

d – common difference between two consecutive elements

a1 – first element in the series

ar – rth element in the series

Common difference (d) = a2 – a1

Geometric Progression

a = a1ri – 1

Sn =a (rn – 1)/(r – 1)

n – number of elements in the series

r – common ratio between two consecutive elements

a1 – first element in the series

ai – ith element in the series

r = a2/a1

Evaluate the series: (a) The given sum, \sum_{k=1}^{20} k       (b) \sum_{j = 1}^{60} 5 j^2   (c)  \sum_{i = 1}^{11} 2 ln3-5i

Solution:

(a) The given sum, \sum_{k=1}^{20} k  can be illustrated as,-

1 + 2 + 3 + …. + 20

The above series is a sum of the first 20 natural numbers and the consecutive numbers have a common difference (d) = 1 and the first number (a1) = 1. Now that the series is in a form of Arithmetic Progression (AP), where, sum of n terms in AP can be given as, 

(S_n) = \frac{n}{2}(2a_1+(n-1)d)

By substituting, n = 20, a1= 1, d=1. 

 S_{20}= \frac{20}{2}(2\times1+(20-1)\times1)=10 \times21 =210      

(b) \sum_{j = 1}^{60} 5 j^2

The above sum is 5 times the sum of squares of the first 60 natural numbers and it can be illustrated as, 

5( 12 + 22 + …. + 602)

The sum of squares for n- terms looks like S_n=1^2+2^2+\dots+n^2   , whose sum can be solved using S_n= \frac{n(n+1)(2n+1)}{6}

So, given sum(S) can be written as,

S = 5 S60

S = 5 \times\frac{(60)(61)(121)}{6} \\ S= 5 \times 10\times61\times121 = 369050

(c)  \sum_{i = 1}^{11} 2 ln3-5 i

The series can be seen as a combination of a constant component and a variable. Such series must be decomposed into constant and variable components to solve the series easily and the above sum can be illustrated as,

2ln3-5(1) +2ln3-5(2)+\dots+2ln3-5(11)

Adding up all the elements,

S = 2ln3\underbrace{(1+1+\dots+1)}_\textit{11 times} - 5(1+2+3+ \dots+11)

Sum of n-terms in AP with common difference d =1 can be written as S = \frac{n(n+1)}{2}   , where n = 11, as in the query.

S= 2ln3(11) - 5 \frac{(11)(12)}{2}

S = 22ln3 – 330 

S = -305.83

Similar Problems 

Question 1: Solve the series, \sum_{r=1}^{10} r^2 + r

Solution :

The above series can be illustrated as follows,

(12+22+ …+102 ) + (1 + 2+… 10) 

This series can be decomposed into two independent series, where one series is the sum of squares of the first 10 natural numbers and the other series would be the sum of the first natural numbers. 

S = S1 + S2

S1 = n(n + 1)(2n + 1)/6 , sum of squares of first n natural numbers

S2 = n(n + 1)/2 ,  sum of first n natural numbers

By substituting n = 10 , S1 = 385 and S2 = 55, so 
S = 385 + 55

S = 440 

Question 2: Solve the series\sum_{r=1}^{20}r^3 + r^2

Solution : 

This series can be decomposed into two independent series just like the above question, where one series is the sum of cubes of the first 20 natural numbers and the other series is the sum of squares of the first 20 natural numbers. Sum (S) can be written as the sum of the independent series.

S = (13 + 23 + 33 + … + 203 ) + (12 + 22 + … + 202)

S = S1 + S2

S1 = n2(n + 1)2/4 , sum of cubes of first n natural numbers

S2 = n(n + 1)(2n + 1)/6 , sum of squares of first n natural numbers

By substituting n = 20, S1 = 44100 and S2 = 2870

As said, S is the composition of S1 and S2, So,

S = S1 + S2 = 46970

Question 3: Solve the series, \sum_{r=0}^{10} 2^r

Solution :

The series can be illustrated as,

S = 1 + 2 + 22 + 23 + … 210 

The above series follows geometric progression with common ratio (r) = 2/1 = 2 , a1 = 1, an = 210, n = 11 elements

S = a1 (rn – 1) / (r – 1) , S -> Sum of elements in GP 

By substituting a1 = 1 , r = 2, n = 11

S = 1 (211 -1) /(2 – 1) 

S = 211 – 1 .


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