Summation of numbers in a sequence

 Write a python program to sum of N natural numbers [even,odd,divisible by n]

So, the immediate way to solve this is by putting it in a loop, use flow control to weed out the unnecessary values.

You can also use range to generate the progression.

But using formula for summation of numbers in a sequence,summation for even large values of N can be done efficiently.

What is a natural number?

0 is usually considered a whole number and excluded from inclusion in the natural numbers.

So numbers 1,2,3,... are considered natural numbers.

Program using a loop

# summation of N natural numbers
N = 25
sum=0
for num in range(N+1):
    sum+=num
print("Sum of first",N,"natural numbers is",sum)

Output

python tyl_naturalNumbers.py
Sum of first 25 natural numbers is 325


Definition of Arithmetic Progression: A sequence of numbers is known as an arithmetic progression (A.P.) if the difference of the term and the preceding term is always same or constant.

General form of arithmetic progression: an = a + (n - 1)d.

Therefore, nth term of an Arithmetic Progress whose first term = ‘a’ and common difference = ‘d’ is a = a + (n - 1)d.


Sum of first n terms in an arithmetic progression.

https://www.math-only-math.com/sum-of-the-first-n-terms-of-an-arithmetic-progression.html

S =n/2 [2a + (n - 1)d]

a=1, d=1

S= n/2[2+n-1] 

S = n/2[n+1]


>>> n=25

>>> for i in range(1,n+1):

...           sum+=i

>>>sum

325

Using the formula

>>> (n/2)*(n+1)

325.0


Sum of first n odd terms in an arithmetic progression.

S =n/2 [2a + (n - 1)d]

a=1, d = 2

S=n/2[2+(n-1)2]

S=n/2[2+2n-2]

S=n/2[2n]

S=n*n

>>> for i in range(1,(n*2)+1,2):

...     sum+=i

...

>>> sum

625


Using the formula

>>> n*n

625


Sum of first n even terms in an arithmetic progression.

S =n/2 [2a + (n - 1)d]

a=2, d=2

S=n/2[4+(n-1)2]

S=n/2[4+2n-2]

S=n/2[2+2n]

S= n[1+n]

S=n*(1+n)

>>>n=25

>>> sum_even=0

>>> for i in range(2, (n*2)+1,2):

...     sum_even+=i

...

>>> sum_even

650


Using formula:

>>>n=25

>>> n*(1+n)

650


Sum of first n divisible by k terms in an arithmetic progression.

S =n/2 [2a + (n - 1)d]

a=k d=k

S=n/2[2k+(n-1)k]

S = nk+[nk(n-1)] / 2

S= nk(1+(n-1)/2)

S=nk[(n+1)/2]


>>> for i in range(k,(n*k)+1,k):

...     sum_divk+=i

...

>>> sum_divk

975


Using the formula:

>>> n*k * ( (n+1) /2 )

975.0




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