1. Prove that 12 + 22 + 32 + ⋯ + 𝑛𝑛2 = 𝑛𝑛(𝑛𝑛+1)6(2𝑛𝑛+1) , ∀𝑛𝑛 ∈ ℕ by using the principle of
mathematical induction.
To prove the given equation using mathematical induction, we need to follow two steps:
Step 1: Show that the equation holds true for the base case (n=1).
Step 2: Assume that the equation holds true for some arbitrary value k and then prove that it also holds true for (k+1).
Let's proceed with the proof:
Step 1: Base Case (n=1)
When n=1, the equation becomes:
12 = 1(1+1)/6(2*1+1)
12 = 1/6(2+1)
12 = 1/6(3)
12 = 1/2
12 = 1
The equation holds true for n=1.
Step 2: Inductive Step
Assume that the equation holds true for some arbitrary value k, i.e.,
12 + 22 + 32 + ⋯ + kk2 = kk(k+1)/6(2k+1)
Now, we need to prove that the equation also holds true for k+1.
We add (k+1)^3 to both sides of the equation:
12 + 22 + 32 + ⋯ + kk2 + (k+1)^3 = kk(k+1)/6(2k+1) + (k+1)^3
Simplifying the right side:
kk(k+1)/6(2k+1) + (k+1)^3
= (kk(k+1) + 6(2k+1)*(k+1)^3)/6(2k+1)
= (k^2(k+1) + 6(k+1)^4)/6(2k+1)
= (k+1)(k^2 + 6(k+1)^3)/6(2k+1)
Now we need to show that this expression is equal to (k+1)((k+1)+1)/6(2(k+1)+1), which is the equation we want to prove.
(k+1)((k+1)+1)/6(2(k+1)+1)
= (k+1)(k+2)/6(2k+3)
To prove that (k+1)(k^2 + 6(k+1)^3)/6(2k+1) = (k+1)(k+2)/6(2k+3), we can cancel out the common factors on the numerator and denominator.
(k+1)(k^2 + 6(k+1)^3)/6(2k+1)
= (k^2 + 6(k+1)^3)/(2k+1)
= [(k+1)(k^2 + 6(k+1)^3)]/[(k+1)(2k+1)]
= (k^2 + 6(k+1)^3)/(2k+1)
We can see that this expression is indeed equal to (k+1)(k+2)/6(2k+3), which means that the equation holds true for (k+1) as well.
By the principle of mathematical induction, we have proven that for all natural numbers n, 12 + 22 + 32 + ⋯ + nn2 = nn(n+1)/6(2n+1).
To prove the equation using the principle of mathematical induction, we need to follow the steps below:
Step 1: Base Case
Show that the equation holds true for the initial value of n. Let's plug in n = 1 and verify if the equation holds:
12 = 1(1 + 1)/6(2*1 + 1)
1 = 1/6(3)
1 = 1/2.
The equation holds true for n = 1.
Step 2: Inductive Hypothesis
Assume that the equation holds true for some positive integer k, where k >= 1.
That is, assume:
12 + 22 + 32 + ... + k^2 = k(k+1)/6(2k+1)
Step 3: Inductive Step
Prove that the equation also holds true for k+1 using the inductive hypothesis.
Adding (k+1)^2 to both sides of the equation:
12 + 22 + 32 + ... + k^2 + (k+1)^2 = k(k+1)/6(2k+1) + (k+1)^2.
Now, we can simplify the right-hand side of the equation:
k(k+1)/6(2k+1) + (k+1)^2
= [(k^2 + k)/6(2k+1)] + (k^2 + 2k + 1)
= (k^2 + k + 6k^2 + 3k + 6)/(6(2k+1))
= (7k^2 + 4k + 6)/(6(2k+1))
= (k+1)(7k+6)/(6(2k+1)).
This expression can be simplified further:
(k+1)(7k+6)/(6(2k+1))
= (k+1)(7k+6)/(12k + 6)
= (k+1)(7k+6)(1)/(12k + 6)
= (k+1)(7k+6)/(2(6k + 3))
= (k+1)(7k+6)/[2 * 3(2k+1)]
= (k+1)(7k+6)/[6(2k+1)]
= (k+1)(7k+6)/6(2k+1)
= (k+1)(k+2)/6(2k+1).
Notice that this expression matches the right-hand side of the original equation for n = k+1, as required.
Hence, the equation holds true for n = k+1, given that it holds for n = k.
By the principle of mathematical induction, we have proved the equation:
12 + 22 + 32 + ... + n^2 = n(n+1)/6(2n+1), for all n ∈ ℕ.
To prove the given statement using the principle of mathematical induction, we need to follow these steps:
Step 1: Base Case
We start by checking if the statement is true for the smallest possible value of n. In this case, n = 1.
When n = 1, the left-hand side (LHS) of the equation becomes:
12 = 1(1+1)6(2(1)+1) = 16.
So, the equation holds true for n = 1.
Step 2: Inductive Hypothesis
Assume that the equation holds true for some positive integer k. This is our Inductive Hypothesis (IH).
That is, assume:
12 + 22 + 32 + ⋯ + k2 = k(k+1)6(2k+1)
Step 3: Inductive Step
We need to prove that the equation holds true for n = k+1 using the assumption made in the inductive hypothesis.
So, we consider the sum of the first (k+1) terms, denoted by S.
S = 12 + 22 + 32 + ⋯ + k2 + (k+1)2
Using the inductive hypothesis, we can replace the sum of the first k squared terms:
S = [k(k+1)6(2k+1)] + (k+1)2
Now, we simplify this expression:
S = [(k+1)(k)6(2k+1)] + (k+1)2
= (k+1)[(k)6(2k+1) + (k+1)]
Expanding further:
S = (k+1)[2k3 + 3k2 + k + k +1]
= (k+1)[2k3 + 3k2 + 2k +1]
Next, we simplify the expression inside the brackets:
S = (k+1)[2k3 + 3k2 + 2k + 1]
= (k+1)[2k3 + 3k2 + 2k + 1 × 1k3]
= (k+1)[2k3 + 3k2 + 2k2 +k3]
= (k+1)[2k3 + k3 + 3k2 + 2k2]
= (k+1)[3k3 + 5k2 + 2k]
Now, we can factor out k from the expression inside the brackets:
S = k(k+1)[3k2 + 5k + 2]
Another factorization:
S = k(k+1)(k+2)(3k+2)
Finally, we can compare this expression with the right-hand side of the given equation:
S = k(k+1)(k+2)(3k+2) = (k+1)(k+1+1)6(2(k+1)+1)
This shows that the statement holds for n = k+1 as well.
Since we have verified the base case and proved the inductive step, we can conclude that the given equation holds for all natural numbers n, using the principle of mathematical induction.