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Series 15
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The program is written here to print the series 2 15 41 80 132 197 275 366 470 587.. For loop is used here to print the series. The value of the first term of the series is 2. The value of the nth term of the series is: (13 × n × (n – 1)) / 2 + 2
For example, if the input is 5,
1st term: (13×1×0)/2+2=2
2nd term: (13×2×1)/2+2=15
3rd term: (13×3×2)/2+2=41
4th term: (13×4×3)/2+2=80
5th term: (13×5×4)/2+2=132
Therefore, the series will be: 2, 15, 41, 80, 132.
Algorithm
INPUT: Number of terms
OUTPUT: The aforesaid series upto nth term
PROCESS:
Step 1: [Taking the input]
Read n [Number of terms]
Step 2: [Printing the series]
Set t<-0
Print "The series is: "
For i=1 to n repeat
Set t<-(13 × i × (i - 1)) / 2 + 2
Print t
[End of ‘for’ loop]
Step 3: Stop.
Code
TIME COMPLEXITY:
for(i=1;i<=n;i++)-----------------------------------O(n)
{
t=(13 * i * (i - 1)) / 2 + 2;
printf("%d ",t);
}
The time complexity of this program is O(n) where ‘n’ is the number of terms of the series.
SPACE COMPLEXITY:
The space complexity of this program is O(1) as it requires a constant number of memory spaces for any given input.
