Given the power series 2 (n + 2) A x(n + 1) n = 0 1. Force the Exponent "n + 1" to be "R" like we did in the last lesson. 2. Now use what we have learned in this lesson to force the new series to start at "R = 3". 00 00 A E (n + 2) A, x(n + 1) 1) A (R - 1) * n = 0 R = 0 00 00 B 2 (n + 2) A x(n +1) E (R + 1) A (R - 1) * n =0 R = 1 E (n + 2) A x(n + 1) = E (R- 1) A R -) (R- 1) n = 0 R = -1 00 00 (D 2 (n + 2) A x(n + 1) Σ (R+2) A n = 0 R = 0 00 E (n + 2) A_ xln + 1) I (R + 3) A E (R +1) n = 0 R = 1 00 E (n + 2) A, xla + 1) Ž (R + 3) Ag x* (F) n = 0 R = 1

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A: To evaluate the given integral

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A: 4y = 12 y = 3

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A:

Q: 3b-12 = 30

A:

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A: Here Given that, 3p-p

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A:

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A: Here we take a general matrix of the given set then find the basis.

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A: For the solution follow the next step.

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A:

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A:

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A:

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Q: I need help with 6b

A: