Let’s consider the series 1, 3, 6, 10, 15, 21……..
First term =1 = ((1 ×2))/2
2^{nd} term = 1+2 = 3 = ((2 ×3))/2
3^{rd} term = 1+2+3 = 6 = ((3 ×4))/2 …………. ………….
n^{th} term= 1+2+3+……..+n = ∑n = n(n+1)/2.
Sum to first “n” natural numbers
1+ 2 + 3 + ………..+ n
∑n = n(n+1)/2
Note: These numbers 1, 3, 6, 10, 15……. can be called triangular numbers because they can be represented graphically in the form of triangles(given below).
1 3 6
Example: What is the 312^{th} term of the series 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5,…………..?
a) 24
b) 22
c) 25
d) 20
Solution:
You are required to define the pattern in the series in variables.
Here 1 is the first term, the next term is 2 till the 3^{rd} term, 3 till the 6^{th} term, 4 till the 10^{th} term, n till the n(n+1)/2 th term.
Thus, 210^{th} term will be = 20 as (20 × 21)/2 = 210.
So, 300^{th} term = 24, because (24 × 25)/2 = 300.
Next 25 terms will be 25. So, the required answer is 25.

Sum of consecutive triangular numbers will be square numbers
What is a square number?? Quite simply, it is a “square”.
Example: 1, 3, 6, 10, 15, 21,……….. is the series of numbers up to n.
This is logical because the sum of two triangles is a square.
9 is a perfect square. Now you graphically know why a square number is called a square.

Difference between squares of triangular numbers will be cube numbers
Example: Sum of squares of first “n” natural numbers
1^{2}+2^{2}+3^{2}+4^{2}+……….n^{2}
∑n²= (n(n+1)×(2n+1))/6
Important Application:
Find the total number of squares in a Chessboard is an 8 × 8 square.
Total number of squares in an n × n square = ∑n^{2}
We can explain this method using the unitary method (i.e. by finding out the pattern using smaller values of n) In a 1×1 square,
Total no.of squares = 1
In a 2 × 2 square there will be 4 squares of area 1 cm² And 1 square of area 4 cm² Total squares = 1 + 4 = 5.
In a 3 × 3 square there will be 9 squares of area 1cm² 4 squares of area 4 cm² and 1 square of area 9cm² Total = 1 + 4 + 9.
By following the pattern, total number of squares in an n × n square = ∑n^{2} = (n(n+1) (2n+1))/6.
∴ A Chessboard is an 8 × 8 square, by substituting n = 8 in the above equation we will get the total number of squares on a chess board as ((8×9×17))/6 = 204.
Illustrations:
1. A person wrote first N consecutive numbers (natural) on a board and found their sum. Another person came and deleted the smallest number and found the sum of remaining numbers. The process continued until N remained on the board.
The average of all the sums is found to out to be 105. Find N?
a) 17
b) 20
c) 19
d) 23
Solution:
1 2 3 4 5……. n
2 3 4 5……. n
3 4 5……. n
4 5……. n
Sum of all terms of the series above = (1) +(2 + 2) +( 3 + 3 + 3)…….(n + n + n…n times)
= 1^{2} + 2^{2} + 3^{2}+……..n^{2}
= ∑n^{2}
Average of all sums = ∑ n^{2}/n = (n(n+1) (2n+1))/6n = ((n+1) (2n+1))/6 = 105
=> (n+1)(2n+1) = 630 => n = 17.
2. Find the sum of the first 20 terms of the series 2 × 3, 4 × 6, 6 × 9, 8 × 12,………..
a) 10710
b) 12654
c) 14880
d) 17220
Solution:
In these kinds of question, the 1^{st} step is to find out the n^{th} term of the series by trial and error ( by following the pattern).
Here, n^{th} term = 2n × 3n
Now take 2n × 3n = 6n^{2} = 6 × (n(n+1).(2n+1))/6 = n ( n+1 ) ( 2n+ 1)
Then put n = 20 to get the required answer as 17220.

Sum of cubes of first “n” natural numbers
1^{3} + 2^{3} + 3^{3} +……………n^{3}
∑n^{3}= (n^{2} (n+1)^{2})/4 = [n(n+1)/2]^{2} = (∑n)^{2}
Important Applications
1. Total number of rectangles in an n × n square
= ∑n^{3}= (n^{2} (n+1)^{2})/4
In an n × n square there will be (n + 1) horizontal lines and (n + 1) vertical lines.
To form a rectangle we need to select 2 horizontal lines from (n + 1) horizontal lines in (n+1)_{C2} ways and 2 vertical lines from(n + 1) vertical lines in (n+1)_{C2} ways.
So the number of rectangles =(n+1)_{C2x} (n+1)_{C2}= (n^{2} (n+1)^{2})/4 = (∑n)^{2} =∑n^{3}
Example: Find the number of rectangles in a chess board A chess board is an 8 × 8 square.
Solution:
Here n=8.
Number of rectangles= ?n^{3}= (n^{2} (n+1)^{2})/4 = (?n)^{2} = [(8 × 9)/2]^{2} = 1296.
2. Fractions in series
Evaluate 1/1×2+1/2×3+1/3×4…..+1/99×100
Each term can be written as 1/1×2= 11/2
1/2×3=1/21/3
.
.
1/99×100=1/99–1/100
Thus the series can be generalised as follows
1/n(n+1) = [1/n–1/(n+1)]
Look at this example,
Find the sum of the series N = 1/3 + 1/15 + 1/35 ………… 100 terms
a) 99/999
b) 1/199
c) 100/201
d) none of these
Solution:
1/3=1/3×1
1/15=1/3×5
1/35=1/((7×5) )…..
Here the difference d = 2,
Thus the sum can be written as ½x[11/3+1/31/5+1/51/7+1/7………………………………………+1/1991/201] (1/2)[11/201]=200/2×201=100/201
It thus follows that
1/n(n+d) = [1/n–1/(n+d)] (in the case above, d=2)
Example: Evaluate (1 – 5/6 + 7/12 – 9/20 + 11/30 – 13/42 + 15/56 – 17/72 + 19/90)
Solution:
Each term can be expressed as the sum of two fractions as follows:
5/6 = 1/2 + 1/3
7/12 = 1/3 + 1/4 . .
19/90 = 1/9 + 1/10
This reduces to 1 – (1/2 + 1/3) + (1/3 + 1/4) – (1/4 + 1/5)……. + (1/9 + 1/10) = 1 – 1/2 + 1/10 = 3/5
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