Question

Find an expression in terms of n for the nth term of this sequence, 15,19,25,33,43

Answer 1

Hello,

Explanation:

P(n)=ax²+bx+c (quadratic sequence)

with a=1,b=1,c=13

`\\\boxed{u_n=n^2+n+13}\\`

Answer 2

• tₙ = n² + n + 13

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Given sequence:

• 15, 19, 25, 33, 43, ...

We can see this not AP or GP as the difference or ratio of consecutive terms is not common.

But we can see the difference between terms has common difference of 2:

• 19 - 15 = 4
• 25 - 19 = 6
• 33 - 25 = 8
• 43 - 33 = 10

and

• 10 - 8 = 2
• 8 - 6 = 2
• 6 - 4 = 2

It means the sequence is of the second degree, in other words it is quadratic.

In general it is of the form:

• tₙ = an² + bn + c, where a, b, c are coefficients, n - is the number of the terms.

Use the terms of the given sequence to work out unknown coefficients.

t₀ = 15 - (4 - 2) = 15 - 2 = 13

• a(0)² + b(0) + c = 13
• c = 13

It gave us the value of c.

t₁ = 15

• a(1)² + b(1) + c = 15
• a + b + c = 15
• a + b + 13 = 15
• a + b = 2

t₂ = 19

• a(2)² + b(2) + c = 19
• 4a + 2b + c = 19
• 4a + 2b + 13 = 19
• 4a + 2b = 6

Compare the last two equations and solve for a and b:

• a = 2 - b

• 4(2 - b) + 2b = 6
• 8 - 4b + 2b = 6
• 2b = 8 - 6
• 2b = 2
• b = 1

Then

• a = 2 - 1
• a = 1

So we know the coefficients:

• a = 1, b = 1, c = 13

The nth term equation is:

• tₙ = n² + n + 13