2(2^1) +3(2^2) + 4(2^3)+...+n^3 = [n(n+1)/2]^2
using the mathematical induction
please help!
The right hand side [n(n+1)/2]^2 applies to the sum Σi³ i=1 to n.
The right hand side [n(n+1)/2]^2 applies to the sum Σi³ i=1 to n.
If the left hand side is correct, then it should read:
2(2^1) +3(2^2) + 4(2^3)+...+(n+1)(2n) = n*2n+1
You can proceed the same way as the answer to the previous post. Feel free to show your work if you have problems.
To prove the given equation using mathematical induction, we need to follow two steps:
Step 1: Base Case
For the base case, we substitute the smallest possible value of n (usually 1) into the equation and check if it holds true.
Let's substitute n = 1 into the equation:
2(2^1) = [1(1+1)/2]^2
4 = [1(2)/2]^2
4 = [1]^2
4 = 4
Since the equation holds true for n = 1, the base case is satisfied.
Step 2: Inductive Step
In the inductive step, we assume that the equation holds true for some arbitrary positive integer k and then prove that it also holds true for k+1.
Assumption: Assume that the equation holds true for k:
2(2^1) + 3(2^2) + 4(2^3) + ... + k^3 = [k(k+1)/2]^2
Now, we need to prove that it holds true for k+1:
2(2^1) + 3(2^2) + 4(2^3) + ... + k^3 + (k+1)^3 = [(k+1)(k+2)/2]^2
To do this, we start with the left side of the equation and try to manipulate it to match the right side. Let's expand the last term:
2(2^1) + 3(2^2) + 4(2^3) + ... + k^3 + (k+1)^3
= [k(k+1)/2]^2 + (k+1)^3
= [(k^2 + k)/2]^2 + (k+1)^3
Next, let's simplify the equation:
[(k^2 + k)/2]^2 + (k+1)^3
= [(k^2 + k)^2/4] + (k+1)^3
= [(k^4 + 2k^3 + k^2)/4] + (k+1)^3
= [k^4 + 2k^3 + k^2 + 4(k+1)^3]/4
= [k^4 + 8k^3 + 20k^2 + 24k + 12]/4
Now, let's work on the right side of the equation using the formula for the sum of cubes:
[(k+1)(k+2)/2]^2
= [(k^2 + 3k + 2)/2]^2
= [(k^2 + 3k + 2)^2]/4
= [k^4 + 6k^3 + 13k^2 + 12k + 4]/4
Therefore, if we can show that both sides of the equation are equal, the induction will be complete.
Comparing the simplified equation on the left side to the equation on the right side, we can see that they match:
[k^4 + 8k^3 + 20k^2 + 24k + 12]/4 = [k^4 + 6k^3 + 13k^2 + 12k + 4]/4
Since the left side is equal to the right side, the equation holds true for k+1 as well.
Step 1 and Step 2 are both complete, so by mathematical induction, we have proven that:
2(2^1) + 3(2^2) + 4(2^3) + ... + n^3 = [n(n+1)/2]^2