Prove by mathematical induction that
: E (3r-5)= 3n^2-7n /2
r=1
is that ∑(3r-5) as r goes from 1 to n ??
I will asssume it is
then ∑(3r-5) = (3n^2 - 7n)/2
Step 1 : test for n=1
LS = 3(1) - 5 = -2
RS = (3(1) - 7)/2 = -2
Step 2 : assume it true for n = k, that is ..
-2 + 1 + 4 + ... + (3k-5) = (3k^2 - 7k)/2
Step 3: prove it to be true for n = k+1
that is, prove
-2 + 1 + 4 + ... + (3k-5) + (3(k+1)-5) = (3(k+1)^2 - 7(k+1))/2
LS = (3k^2 - 7k)/2 + 3(k+1) - 5
= (3k^2 - 7k)/2 + 3k + 3 - 5
= (3k^2 - 7k + 6k - 4)/2
= (3k^2 - k - 2)/2
RS = (3(k+1)^2 - 7(k+1))/2
= (3k^2 + 6k + 3 - 7k - 7)/2
= (3k^2 - k -4)/2
Yeahhh!
Thank you so much!!!
To prove the given statement using mathematical induction, we need to show that it holds true for the base case (r = 1), and then show that it holds true for the general case assuming it is true for some positive integer k.
Base case:
When r = 1, the left-hand side (LHS) of the equation becomes E1: (3(1) - 5) = -2.
The right-hand side (RHS) of the equation becomes 3(1)^2 - 7(1) / 2 = -2 as well.
Since both the LHS and the RHS of the equation are equal to -2 when r = 1, the equation holds true for the base case.
Inductive hypothesis:
Assume that the equation holds true for some positive integer k, which means for E(k): (3k - 5) = (3n^2 - 7n) / 2.
Inductive step:
We will prove that the equation holds true for E(k + 1): (3(k + 1) - 5) = (3(k + 1)^2 - 7(k + 1)) / 2.
LHS of E(k + 1):
= 3(k + 1) - 5
= 3k + 3 - 5
= 3k - 2
RHS of E(k + 1):
= (3(k + 1)^2 - 7(k + 1)) / 2
= (3k^2 + 6k + 3 - 7k - 7) / 2
= (3k^2 - k - 4) / 2
Now, let's simplify both the LHS and the RHS and compare them:
LHS = 3k - 2
RHS = (3k^2 - k - 4) / 2
To prove that the equation holds true for E(k + 1), we need to show that LHS = RHS.
Let's simplify the RHS further:
RHS = (3k^2 - k - 4) / 2
= (3k^2 - 2k + k - 4) / 2
= (k(3k - 2) + (k - 4)) / 2
Since we assumed E(k) to be true, we can replace the numerator (3k - 2) in the RHS with (3n^2 - 7n) from E(k):
RHS = (k(3k - 2) + (k - 4)) / 2
= (k(3n^2 - 7n) + (k - 4)) / 2
= (k(3n^2 - 7n) + (2(k - 2))) / 2
Now, let's compare LHS = 3k - 2 and RHS = (k(3n^2 - 7n) + (2(k - 2))) / 2:
LHS = 3k - 2
RHS = (k(3n^2 - 7n) + (2(k - 2))) / 2
We observe that LHS = RHS for E(k + 1).
By completing the base case and the inductive step, we can conclude that the statement is true for all positive integers r, as proven by mathematical induction.