what is the sum of first n odd natural numbers? With detail explanation
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Answered by
1
Answer:
n²
Step-by-step explanation:
First odd number = 1
2nd odd number = 3
3rd odd number = 5
first term = 1
com. dif = 3 - 1 = 2
Therefore,
S = (n/2)[ 2a + (n - 1)d]
= (n/2)[ 2(1) + (n - 1)2]
= (n/2)[ 2 + 2n - 2]
= (n/2)( 2n )
= n²
Answered by
1
n²
Step-by-step explanation:
given:
the first odd number is 1
second is 3
the third is 5 and so on...
here common difference is t2-t1
d = 2
therefore
S = n/2 [2a +(n -1) ×d]
S = n/2 [2×1 +(n-1)×2]
S = n/2 [2 + 2n - 2]
S = n/2 [ 2n]
S = n × n
S= n²
hope it's helpful for you
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