Prove by mathematical induction that it's true for all positive integers with 10n=5n(5+1)

When n = 1, left side = 10, right side = 5(2) = 10 
When n = 2, left side = 10+30 = 40, right side = 10(3) = 30 
So your statement is not true 

Perhaps you mean 10 + 20 + 30 + ... + 10n = 5n(n+1) 

When n = 1, left side = 10, right side = 5(2) = 10 
When n = 2, left side = 10+20 = 30, right side = 10(3) = 30 
That seems better. 

So we've shown base case, now we'll do the inductive step. 

Assume statement is true for n: 10 + 20 + 30 + ... + 10n = 5n(n+1) 
We must show that statement is also true for (n+1), i.e. we must show that 
10 + 20 + 30 + ... + 10(n+1) = 5(n+1)(n+2) 

10 + 20 + 30 + ... + 10(n+1) = (10 + 20 + 30 + ... + 10n) + 10(n+1) 
. . . . . . . . . . . . . . . . . . . . . = 5n(n+1) + 10(n+1) 
. . . . . . . . . . . . . . . . . . . . . = 5(n+1) (n+2) 

So when statement is true for n, it is also true for n+1 

Therefore, by mathematical induction, statement is true for all positive integers n 

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