Question

1+3+5+7+9+11+13+....... find the sum of n terms

Answer 1

here, d= 2 ,a = 1
So , Sn = n/2[a+(n-1)d]
=> Sn = n/2 [1+(n-1)2]
=> Sn = n/2 [1+2n-2 ]
=> Sn = n/2 [2n-1]
=> Sn = n^2 -n/2
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