Using Mathematical Induction, prove the following:
a) 2+4+6+.....+2n= n(n+1)
To prove this statement using mathematical induction, we follow these steps:
Step 1: Base Case
We start by checking the base case, which is typically when n = 1 or n = 0. In this case, let's check for n = 1:
The left-hand side (LHS) of the equation is 2 + 4 = 6.
The right-hand side (RHS) of the equation is 1(1+1) = 1(2) = 2.
Since LHS and RHS are not equal for n = 1, this equation is not true for the base case.
Step 2: Assume the Statement is True for n = k
Next, we assume that the statement is true for some arbitrary but fixed value k. In other words, we assume that:
2 + 4 + 6 + ... + 2k = k(k+1) is true.
Step 3: Prove the Statement is True for n = k+1
Now, we need to prove that if the statement is true for n = k, it is also true for n = k+1. Substituting k+1 into the equation, we have:
2 + 4 + 6 + ... + 2k + 2(k+1)
We can rewrite this as:
k(k+1) + 2(k+1)
Now, we can simplify the expression:
k(k+1) + 2(k+1) = (k+1)(k+2)
This is equal to the right-hand side (RHS) of the equation, which is (k+1)(k+2).
Step 4: Conclusion
Since we have shown that if the statement is true for n = k, it is also true for n = k+1, we can conclude that the statement is true for all positive integers n.
However, in this case, the base case (n = 1) was not true, so the statement 2 + 4 + 6 + ... + 2n = n(n+1) is not true for all positive integers n.
surely you have seen how to show that
1+2+3+...+n = n(n+1)/2
your sum is just twice that.