Prove that m^r can be expressed as sum of m consecutive odd integers. Find the first odd integer in case of 8^5. m and r are natural numbers where r1.

Question

Prove that m^r can be expressed as sum of m consecutive odd integers. Find the first odd integer in case of 8^5. m and r are natural numbers where r>1.

mycroft holmes · Answer

m r = (2a-1)+....+(2b-1) = b 2 - (a-1) 2 = [b-(a-1)][b+a-1] = m r We are given that b-(a-1) = m which means b+(a-1) = m r-1 So choose a such that a = m(m r-2 -1)/2 + 1 and b = m(m r-2 +1)/2 In case of 8 5 it is 2045

GAGANDEEP SINGH · Answer

Yeah solved it a little later. Thanks for sparing for valuable time on this. But your answer to 8^5 is wrong. This was a question in Sequences and Series.

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Prove that m^r can be expressed as sum of m consecutive odd integers. Find the first odd integer in case of 8^5. m and r are natural numbers where r>1.

Prove that m^r can be expressed as sum of m consecutive odd integers. Find the first odd integer in case of 8^5. m and r are natural numbers where r>1.

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Prove that m^r can be expressed as sum of m consecutive odd integers. Find the first odd integer in case of 8^5. m and r are natural numbers where r>1.

Prove that m^r can be expressed as sum of m consecutive odd integers. Find the first odd integer in case of 8^5. m and r are natural numbers where r>1.

2 Answers

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