sum of odd number from 1 to 100 with formula
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Using Arithmetic Progression we can find the sum of odd numbers from 1 to infinity. As we know, the odd_numbers are the numbers that are not divisible by 2. They are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19 and so on. Now, we have to search the sum of these numbers.
Let Sn be the sum of first n odd numbers.
Or, Sn = 1 + 3 + 5 + 7 + 9 +…………………..+ (2n - 1) (Equation 1)
With the help of Arithmetic_Progression(Ar) here, we know, for any series, the sum of numbers is given by;
or, Sn = ½ × n2a+(n−1)d
……..(Equation 2)
Where,
n = number of digits present in the series
a = It is termed as First term of an A.P
d= it is the Common difference in an A.P
Therefore, if we put the values in equation 2 with respect to equation 1, we will get
a = 1, d = 2
Let, last term be, l = (2n - 1)
or, Sn = (n/2) × (a + l)
or, Sn = (n/2) × (1 + 2n – 1)
or, Sn = (n/2) × (2n) = n2
Let Sn be the sum of first n odd numbers.
Or, Sn = 1 + 3 + 5 + 7 + 9 +…………………..+ (2n - 1) (Equation 1)
With the help of Arithmetic_Progression(Ar) here, we know, for any series, the sum of numbers is given by;
or, Sn = ½ × n2a+(n−1)d
……..(Equation 2)
Where,
n = number of digits present in the series
a = It is termed as First term of an A.P
d= it is the Common difference in an A.P
Therefore, if we put the values in equation 2 with respect to equation 1, we will get
a = 1, d = 2
Let, last term be, l = (2n - 1)
or, Sn = (n/2) × (a + l)
or, Sn = (n/2) × (1 + 2n – 1)
or, Sn = (n/2) × (2n) = n2
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Answer:
Solution: As per the list of odd numbers 1 to 100, the largest odd number is 99 and the smallest odd number is 1. So, the required sum is 99+1 = 100.
Step-by-step explanation:
hope it's helps you.
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