How do you find the explicit formula and calculate term 20 for -1, 6, 25, 62, 123?

Answer 1

Look at sequences of differences to construct a formula and find the #20#th term is #7998#.

Each of the last 3 terms looks close to a cube, so I could guess the formula as #a_n = n^3-2#, but let's pretend I didn't spot that...

Put the first sequence in writing:

#color(blue)(-1), 6, 25, 62, 123#

List the differences in that sequence in writing:

#color(blue)(7), 19, 37, 61#

List the differences in that sequence in writing:

#color(blue)(12), 18, 24#

List the differences in that sequence in writing:

#color(blue)(6), 6#
Having reached a constant sequence, we can now use the first number of each sequence to write out a formula for #a_n# as follows:
#a_n = color(blue)(-1)/(0!) +color(blue)(7)/(1!) (n-1) + color(blue)(12)/(2!) (n-1)(n-2) + color(blue)(6)/(3!) (n-1)(n-2)(n-3)#
#=-1+7(n-1)+6(n-1)(n-2)+(n-1)(n-2)(n-3)#
#=-1+7n-7+6n^2-18n+12+n^3-6n^2+11n-6#
#=n^3-2#
So #a_20 = 20^3-2 = 8000-2 = 7998#
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Answer 2

The given sequence is: -1, 6, 25, 62, 123.

To find the explicit formula, we can look for patterns in the differences between consecutive terms.

1st difference: 6 - (-1) = 7 2nd difference: 25 - 6 = 19 3rd difference: 62 - 25 = 37 4th difference: 123 - 62 = 61

We notice that the differences are not constant, but they seem to follow a pattern. The differences themselves form a sequence of odd numbers (7, 19, 37, 61), which are consecutive odd integers starting from 7.

This suggests that the original sequence might be generated by a cubic function, as the differences between consecutive terms are quadratic.

Let's assume the explicit formula is of the form f(n)=an3+bn2+cn+d f(n) = an^3 + bn^2 + cn + d .

Given that the first term is -1, we can find the values of a, b, c, and d by plugging in the values of n and the corresponding terms:

When n=1 n = 1 : a(1)3+b(1)2+c(1)+d=1a(1)^3 + b(1)^2 + c(1) + d = -1 a+b+c+d=1a + b + c + d = -1 ...(1)

When n=2 n = 2 : a(2)3+b(2)2+c(2)+d=6a(2)^3 + b(2)^2 + c(2) + d = 6 8a+4b+2c+d=68a + 4b + 2c + d = 6 ...(2)

When n=3 n = 3 : a(3)3+b(3)2+c(3)+d=25a(3)^3 + b(3)^2 + c(3) + d = 25 27a+9b+3c+d=2527a + 9b + 3c + d = 25 ...(3)

When n=4 n = 4 : a(4)3+b(4)2+c(4)+d=62a(4)^3 + b(4)^2 + c(4) + d = 62 64a+16b+4c+d=6264a + 16b + 4c + d = 62 ...(4)

Solving the system of equations (1)-(4) will give us the values of a, b, c, and d.

Once we have the explicit formula, we can calculate the 20th term by substituting n=20 n = 20 into the formula and evaluating it.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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