1(2)+2(3)+3(4)+...+n(n+1)= n(n+1)(n+2)/2
using the mathematical induction.
please help..
1(2)+2(3)+3(4)+...+n(n+1)= n(n+1)(n+2)/3
(Note: The denominator is 3, not 2)
Basis: n=1
1(2)=2
1(2)(3)/3=2
n=1 is valid.
Assume:
1(2)+2(3)+3(4)+...+k(k+1)= k(k+1)(k+2)/3
is valid for k=n
then try to prove that the relation is valid for k=n+1.
1(2)+2(3)+3(4)+...+n(n+1) + (n+1)(n+2)
= n(n+1)(n+2)/3 + (n+1)(n+2)
= ( n(n+1)(n+2)+3(n+1)(n+2) )/3
= ( n^3+6*n^2+11*n+6 )/3
= (n+1)(n+2)(n+3)/3
= (n+1)*(n+1 +1)*(n+1 +2)/3
which means that the expression is valid also for n+1. QED
To prove the given equality using mathematical induction, we follow these steps:
Step 1: Base case
First, we need to verify if the equation is true for the initial value of n. In this case, n=1:
1(2) = 1(1+1)(1+2)/2
2 = 1(2)(3)/2
2 = 6/2
2 = 2
The equation holds true for n=1.
Step 2: Assume the equation holds for a general case
Assume that the given equation holds true for some positive integer k. This is called the induction hypothesis. We assume:
1(2) + 2(3) + 3(4) + ... + k(k+1) = k(k+1)(k+2)/2
Step 3: Prove the equation holds for the next case
We need to prove that the equation also holds for the next positive integer k+1. So, we add (k+1)(k+2) to both sides of the equation:
1(2) + 2(3) + 3(4) + ... + k(k+1) + (k+1)(k+2) = k(k+1)(k+2)/2 + (k+1)(k+2)
Now, let's manipulate the right-hand side of the equation:
= (k+1)(k+2)/2 * (k+1) + (k+1)(k+2)
= (k+1)[(k+2)/2 * (k+1) + (k+2)]
Now, factor out (k+1)(k+2) from the bracket:
= (k+1)(k+2)[(k+1)/2 + 1]
= (k+1)(k+2)[(k+1 + 2)/2]
= (k+1)(k+2)(k+3)/2
Thus, we have shown that if the equation holds for k, then it also holds for k+1.
Step 4: Conclusion
By proving the base case (n=1) and demonstrating that the equation holds true for k+1 assuming it holds true for k, we have proven the original equation using mathematical induction.