Question

The first three terms of an arithmetic sequence are u1, 5u1-8, and 3u1+8. U1 is equal to 4. Prove by induction that the sum of the first n terms of the sequence is a square number.

Answer 1

Answer: To prove that the sum of the first n terms of the sequence is a square number, we will use mathematical induction.

Base case: When n = 1, the sum of the first term of the sequence is u1 = 4, which is a square number (2^2). So the statement is true for n = 1.

Inductive step: Assume that the statement is true for n = k, which means that the sum of the first k terms of the sequence is a square number. We need to prove that the statement is also true for n = k + 1.

The sum of the first k+1 terms of the sequence is:

S(k+1) = u1 + u2 + u3 + ... + uk + uk+1

We know that the first three terms of the sequence are u1, 5u1-8, and 3u1+8. So we can write:

u2 = 5u1 - 8

u3 = 3u1 + 8

u4 = u3 + d = 3u1 + 8 + d

where d is the common difference of the sequence.

To find the value of d, we can use the formula:

d = u2 - u1 = (5u1 - 8) - u1 = 4u1 - 8

So we have:

u4 = 3u1 + 4u1 - 8 + 8 = 7u1

Now we can write:

S(k+1) = u1 + u2 + u3 + ... + uk + uk+1

S(k+1) = S(k) + uk+1

S(k+1) = n^2 + 7u1 (by the inductive hypothesis)

We need to show that S(k+1) is also a square number. Let's write S(k+1) as:

S(k+1) = n^2 + 7u1 = (n^2 + 2n + 1) + (4u1 - 1)

We can rewrite this as:

S(k+1) = (n+1)^2 + (2u1 - 1)^2

Since both (n+1)^2 and (2u1 - 1)^2 are square numbers, their sum is also a square number. Therefore, S(k+1) is a square number.

Step-by-step explanation: Have a good day:)

Answer 2

Answer:

$S_n=(2n)^2$

Step-by-step explanation:

The first three terms of an arithmetic sequence are:

• $u_1$
• $5u_1-8$
• $3u_1+8$

We are told that u₁ = 4.

Substituting u₁ = 4 into the expressions for the first three terms gives:

• $u_1=4$
• $5u_1-8=5(4)-8=12$
• $3u_1+8=3(4)+8=20$

Therefore, the first three terms of the arithmetic sequence are:

• 4, 12, 20.

The common difference (d), of an arithmetic sequence is the constant difference between consecutive terms.

$12-4=8$

$20-12=8$

Therefore, the common difference of the given sequence is d = 8.

The first term is 4, so a = 4.

The formula for the sum of the first n terms of an arithmetic sequence is:

$\boxed{\begin{minipage}{7.3 cm}\underline{Sum of the first n terms of an arithmetic series}\\\\ S_n=\dfrac{1}{2}n[2a+(n-1)d] \\\\where:\\\phantom{ww} \bullet a is the first term. \\ \phantom{ww} \bullet d is the common difference.\\ \phantom{ww} \bullet n is the position of the term.\\\end{minipage}}$

Substitute a = 4 and d = 8 into the equation:

$S_n=\dfrac{1}{2}n\left[2(4)+(n-1)8\right]$

Simplify:

$S_n=\dfrac{1}{2}n\left[8+8n-8\right]$

$S_n=\dfrac{1}{2}n\left[8n\right]$

$S_n=4n^2$

$S_n=2^2 \cdot n^2$

$S_n=(2n)^2$

Therefore, the sum of the first n terms of the given arithmetic sequence is (2n)², where n is the position of the term. Hence proving that that Sₙ is a square number.