X4 -110.) If x2 = k + 1, thenx2 + 1(A) k(B) k2(C) k + 2(D) k-2

1.	X4 -110.) If x2 = k + 1, thenx2 + 1(A) k(B) k2(C) k + 2(D) k-2
Answer» Answer: PROBLEM 1. In each part determine if the series is convergent or divergent. If ... 5/8. 1 − 1/2. = 2. (5. 8. ) = 5. 4. B. ∞. ∑ k=1. 3ek. Answer: We have. ∞. ∑ k=1 ... D. ∞. ∑ k=3. 1 k(k + 1) . Answer: Partial fractions GIVES. 1 k(k + 1). = 1 k. −. 1 k + 1 ... )k . Then we have k. √ AK = [( k. 2k + 1. )k]1/k. = k. 2k + 1. = 1. 2+1/k. →. 1. 2+0

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Answer:

PROBLEM 1. In each part determine if the series is convergent or divergent. If ... 5/8. 1 − 1/2. = 2. (5. 8. ) = 5. 4. B. ∞. ∑ k=1. 3ek. Answer: We have. ∞. ∑ k=1 ... D. ∞. ∑ k=3. 1 k(k + 1) . Answer: Partial fractions GIVES. 1 k(k + 1). = 1 k. −. 1 k + 1 ... )k . Then we have k. √ AK = [( k. 2k + 1. )k]1/k. = k. 2k + 1. = 1. 2+1/k. →. 1. 2+0

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X4 -110.) If x2 = k + 1, thenx2 + 1(A) k(B) k2(C) k + 2(D) k-2​

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X4 -110.) If x2 = k + 1, thenx2 + 1(A) k(B) k2(C) k + 2(D) k-2