find the sum of the sequences of first 50 natural numbers
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Answers
Step-by-step explanation:
Sum of 50 natural numbers
Since the problem statement is asking for the sum of the first 50 natural numbers and that is a large amount to calculate, and also the first 50 natural numbers is actually an AP with a common difference of 1, finding the generalized formula for this would be better. In order to find the sum of more than one natural number, let’s take the sum of the first n natural number. Using the formula earlier discussed in the article we we will find the first n natural number
T(n) = 1+ 2+ 3+ … + n
add T(n) on both sides
⇒T(n)+T(n) = 1 + 2 + 3 + … + n +T(n)
⇒T(n)+T(n) = 1 + 2 + 3 + … + (n-1) + n + n + (n-1) + (n-2) + … + 2 + 1
Now, pair of the the terms such that there sum equal to be (n+1)2T(n)= (1 + n) + (2 + n-1) + (3 + n-2) + … + (n-1 + 2) + (n + 1)
All the n pair-sums are equal to (n+1),
⇒2T(n)= (n+1) + (n+1) + (n+1) + … + (n+1) + (n+1)
⇒2T(n) = n (n+1)
⇒T(n) = n (n+1) /2
So the formula to deduce the sum of the first n natural number. So let calculate the sum of the first 50 Natural Numbers is written as follows,T(50) = 50(50+1)/2
T(50)=25×51
T(50)=1275
And hence the sum of the first 50 natural numbers to be 1275.
Similar Problems
Question 1: What is the difference between twenty and ten natural numbers?
Solution:
Lets first calculate the sum of the first of ten natural numbers by using the formula
T(n) = n (n+1) /2
Therefrore n=10,
⇒ T(10) = 10(10+1)/2
⇒ T(10) = (10×11)/2
⇒ T(10) = 110/2
⇒ T(10) = 55
Now for n = 20,
⇒ T(20) = 20(20+1)/2
⇒ T(20) = (20×21)/2
⇒ T(20) = 420/2
⇒ T(20) = 210
Therefore, T(20) -T(10) = 210-55 = 155