By proving,prove 1+4+7....+(3n-2) = n(3n-1)/2 using mathematical induction for n > or = 1
To prove the statement using mathematical induction, we need to follow these steps:
Step 1: Base case: First, we need to prove that the formula holds for the smallest valid value of n, which is n = 1.
For n = 1:
1(3(1) - 1) / 2 = 1(3 - 1) / 2 = 1(2) / 2 = 1
Which is equal to the left-hand side of our equation:
1 = 1
Step 2: Inductive hypothesis: Assume that the formula holds for some arbitrary positive integer k ≥ 1. That is,
1 + 4 + 7 + ... + (3k-2) = k(3k-1) / 2
Step 3: Inductive step: Now we need to prove that if the formula holds for k, then it also holds for k + 1.
We can express the sum from 1 to (3k+1) terms as the sum from 1 to (3k-2) terms plus (3k+1):
1 + 4 + 7 + ... + (3k-2) + (3k+1)
By the inductive hypothesis, we can substitute the sum from 1 to (3k-2) terms with k(3k-1)/2:
k(3k-1) / 2 + (3k+1)
Simplifying the expression:
(3k^2 - k + 6k + 2) / 2
(3k^2 + 5k + 2) / 2
Factoring the numerator:
[(3k+1)(k+2)] / 2
Now, we can rewrite the expression in terms of (k+1):
(k+1)(3(k+1) - 1) / 2
(k+1)(3k+2) / 2
This is equal to the right-hand side of our equation.
Step 4: Conclusion: Since we have shown that the equation holds for n=1, and we have proven that if it holds for k, then it holds for k + 1, we can conclude that the equation holds for all positive integers n ≥ 1 by mathematical induction.
Therefore, we have proved that 1 + 4 + 7 + ... + (3n-2) = n(3n-1)/2 for n ≥ 1 by mathematical induction.
To prove the given statement using mathematical induction, we will follow two steps:
Step 1: Base Case
Step 2: Inductive Step
Step 1: Base Case
Let's check the statement for the base case when n = 1.
For n = 1, the left-hand side (LHS) of the equation becomes:
1 + 4(1) - 2 = 3
The right-hand side (RHS) becomes:
1(3(1) - 1)/2 = 1(2)/2 = 1
Since LHS = RHS for n = 1, the statement holds true for the base case.
Step 2: Inductive Step
Assume the statement holds true for some arbitrary positive integer k (inductive hypothesis) i.e., assuming that:
1 + 4 + 7 + ... + (3k - 2) = k(3k - 1)/2
Now, we need to prove the statement holds true for the next positive integer k + 1.
Consider the sum of the first (k + 1) terms:
1 + 4 + 7 + ... + (3k - 2) + (3(k + 1) - 2)
Rearranging the terms, we get:
[1 + 4 + 7 + ... + (3k - 2)] + (3(k + 1) - 2)
Using the inductive hypothesis, we can substitute k(3k - 1)/2 for the first part:
k(3k - 1)/2 + (3(k + 1) - 2)
Simplifying, we have:
k(3k - 1)/2 + 3(k + 1) - 2
Expanding and simplifying further:
(k(3k - 1) + 6(k + 1) - 4)/2
Rearranging and grouping like terms:
(3k^2 + 5k + 2)/2
Factoring the numerator:
[(3k + 2)(k + 1)]/2
Simplifying the expression:
(k + 1)(3k + 2)/2
Since this is equivalent to the right-hand side of the statement n(3n - 1)/2 when n = k + 1, the statement holds true for (k + 1).
By verifying the base case and proving the inductive step, we have shown that the statement is true for all positive integers n ≥ 1 using mathematical induction.
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