1+2+3+.....+n<1/8(2n+1)²prove by mathematical induction.
Yippeeee!
This is a two step process 1) show that the left hand side is less than the right hand side when n = 1, then step 2) assume true for n=k and show true for n= k + 1
I'm not able to prove n=k+1
To prove the inequality 1 + 2 + 3 + ... + n < 1/8(2n + 1)² using mathematical induction, we need to do the following steps:
Step 1: Show that the inequality is true for the base case, which is usually n = 1.
Step 2: Assume that the inequality holds true for some arbitrary value k, which is referred to as the "inductive hypothesis".
Step 3: Use the assumption from step 2 to prove that the inequality is true for the next case, k + 1.
Now let's go through these steps in detail.
Step 1: Base Case (n = 1)
We need to prove that the inequality holds true for n = 1. When n = 1, the left side of the inequality becomes 1, and the right side becomes 1/8(2 + 1)² = 1/8(3)² = 1/8(9) = 9/8. Since 1 < 9/8, the inequality is true for n = 1.
Step 2: Inductive Hypothesis
Assume that the inequality holds true for some arbitrary value k. In other words, let's assume that 1 + 2 + 3 + ... + k < 1/8(2k + 1)².
Step 3: Inductive Step (n = k + 1)
Now we need to prove that the inequality is true for n = k + 1 using the assumption from step 2.
Starting with the left side of the inequality:
1 + 2 + 3 + ... + k + (k + 1) = [1 + 2 + 3 + ... + k] + (k + 1)
By the inductive hypothesis, we assumed that 1 + 2 + 3 + ... + k < 1/8(2k + 1)². Substituting this in, we have:
[1/8(2k + 1)²] + (k + 1)
Now let's simplify the right side of the inequality:
1/8(2k + 1)² + (k + 1)
Expanding the square term:
1/8(4k² + 4k + 1) + (k + 1)
Simplifying further:
(4k² + 4k + 1)/8 + (k + 1)
Combining the terms with a common denominator:
(4k² + 4k + 1 + 8k + 8) / 8
Simplifying the numerator:
(4k² + 12k + 9) / 8
Factoring the numerator:
((2k + 3)(2k + 3)) / 8
Simplifying further:
(2k + 3)² / 8 = 1/8(2k + 3)²
Now we have the right side of the inequality in the same form as the inequality we assumed for k. Thus, we can conclude that 1 + 2 + 3 + ... + k + (k + 1) < 1/8(2k + 3)².
Therefore, by mathematical induction, we have proven that 1 + 2 + 3 + ... + n < 1/8(2n + 1)² for all positive integers n.