The sum of the first n odd natural numbers is equal to n2

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poleglit3 · Accepted Answer

Step 1&amp;nbsp;Let 1, 3, 5, 7, 9, n odd numbers be the first.It is an Arithmatic Progression, whose first term&amp;nbsp;(a)=1, and common difference (d) is 2.&amp;nbsp;Step 2&amp;nbsp;The total of these n terms&amp;nbsp;A⋅P=n2[2a+(n−1)d]&amp;nbsp;first n odd numbers added together&amp;nbsp;=n2[2×1+(n−1)×2]&amp;nbsp;=n2[2+2n−2]&amp;nbsp;=n2×n=n2&amp;nbsp;Hence, the sum of first n odd natural numbers is equal to&amp;nbsp;n2. Hence, proved.

nghodlokl · Accepted Answer

S=1+3+5+…+(2n−5)+(2n−3)+(2n−1)  is the sum of the first n odd numbers    We can also write  S=(2n−1)+(2n−3)+(2n−5)+…+5+3+1  Adding both results in sequence  2S=[1+(2n−1]+[3+(2n−3)]+[5+(2n−5)]+…+[(2n−5)+5]+[(2n−3)+3]+[(2n−1)+1]  ⇒2S=(2n)+(2n)+(2n)+…+(2n)+(2n)+(2n) n terms  ⇒2S=(2n)(n)  ⇒2S=2n2  ⇒S=n2  ⇒1+3+5+…+(2n−5)+(2n−3)+(2n−1)=n2

user_27qwe · Accepted Answer

Proving by example Sum of first 3 odd numbers is 32=9 First 3 odd numbers are 1+3+5=9 Hence ,Sum of first n odd natural numbers is n2

The sum of the first n odd natural numbers is

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