for n=1,
1 = 1(2)/2✅
Now assume it is true for n=k
Consider n=k+1
1+4+7+...+(3k-2)+(3(k+1)-2) = k(3k-1)/2 + (3(k+1)-2)
= (3k^2-k+2(3k+3-2))/2
= (3k^2-k+6x+2)/2
= (3k^2+5k+2)/2
= (k+1)(3k+2)/2
= (k+1)(3(k+1)-1)/2
So it is also true for n=k+1
Thus for all n >= 1
By proving,prove 1+4+7....+(3n-2) = n(3n-1)/2 using mathematical induction for n > or = 1
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