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Question Number 185229 by Mingma last updated on 18/Jan/23

Answered by Rasheed.Sindhi last updated on 21/Jan/23

A_S =(n/2)[2c+2(n−1)] G_S =((c(1−2^n ))/(1−2)) 2∙A_S =G_S 2((n/2)[2c+2(n−1)])=((c(1−2^n ))/(1−2)) 2cn+2n(n−1)=c(2^n −1) c(2^n −1)−2cn=2n(n−1) c(2^n −2n−1)=2n(n−1) c=((2n(n−1))/(2^n −2n−1)) For c to be +ve integer 2^n −2n−1 ∣ 2n(n−1) n=1:c=((2(1)(1−1))/(2^1 −2(1)−1))=0∉Z^+ n=2: c=((2(2)(2−1))/(2^2 −2(2)−1))=(4/(−1))=−4∉Z^+ n=3:c=((2(3)(3−1))/(2^3 −2(3)−1))=((12)/(8−6−1))=12✓ n=4:c=((2(4)(4−1))/(2^4 −2(4)−1))=((24)/7)∉Z^+ n=5:c=((2(5)(5−1))/(2^5 −2(5)−1))=((40)/(21))∉Z^+ n=6:c=((2(6)(6−1))/(2^6 −2(6)−1))=((60)/(51))∉Z^+ n>6: 2^n −2n−1>2n(n−1) and hence 2^n −2n−1 ∤ 2n(n−1) c=12 when n=3 Verification: 2(12+14+16)=^(?) 12+24+48 84=84

AS=n2[2c+2(n−1)]GS=c(1−2n)1−22⋅AS=GS2(n2[2c+2(n−1)])=c(1−2n)1−22cn+2n(n−1)=c(2n−1)c(2n−1)−2cn=2n(n−1)c(2n−2n−1)=2n(n−1)c=2n(n−1)2n−2n−1Forctobe+veinteger2n−2n−1∣2n(n−1)n=1:c=2(1)(1−1)21−2(1)−1=0∉Z+n=2:c=2(2)(2−1)22−2(2)−1=4−1=−4∉Z+n=3:c=2(3)(3−1)23−2(3)−1=128−6−1=12✓n=4:c=2(4)(4−1)24−2(4)−1=247∉Z+n=5:c=2(5)(5−1)25−2(5)−1=4021∉Z+n=6:c=2(6)(6−1)26−2(6)−1=6051∉Z+n>6:2n−2n−1>2n(n−1)andhence2n−2n−1∤2n(n−1)c=12whenn=3Verification:2(12+14+16)=?12+24+4884=84

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