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jenyasd209 [6]
4 years ago
6

The recursive formula to describe a sequence is shown below

Mathematics
1 answer:
andreyandreev [35.5K]4 years ago
4 0

Given that the recursive formula for a sequence is a_n=a_{n-1}+2n

The first term of the sequence is a_1=4

We need to determine the first four terms of the sequence.

<u>Second term:</u>

The second term of the sequence can be determined by substituting n = 2 in the recursive formula.

Thus, we have;

a_2=a_{2-1}+2(2)

a_2=a_{1}+2(2)

a_2=4+4

a_2=8

Thus, the second term of the sequence is 8.

<u>Third term:</u>

The third term of the sequence can be determined by substituting n = 3 in the recursive formula.

Thus, we have;

a_3=a_{3-1}+2(3)

a_3=a_{2}+2(3)

a_3=8+6

a_3=14

Thus, the third term of the sequence is 14.

<u>Fourth term:</u>

The fourth term of the sequence can be determined by substituting n = 4 in the recursive formula.

Thus, we have;

a_4=a_{4-1}+2(4)

a_4=a_{3}+2(4)

a_4=14+8

a_4=22

Thus, the fourth term of the sequence is 22.

Hence, the first four terms of the sequence is 4, 8, 14, 22.

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has characteristic equation

<em>r</em> ² + 4<em>r</em> + 4 = (<em>r</em> + 2)² = 0

with a double root at <em>r</em> = -2, so the characteristic solution is

y_c = C_1e^{-2t} + C_2te^{-2t}

For the particular solution corresponding to 12te^{-2t}, we might first try the <em>ansatz</em>

y_p = (At+B)e^{-2t}

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y_p = (At^3+Bt^2)e^{-2t}

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Substituting these into the left side of the ODE gives

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Then the general solution to the ODE is

y = C_1e^{-2t} + C_2te^{-2t} + 2t^3e^{-2t} - 2t - 1

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and so the particular solution satisfying these conditions is

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