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2n次方=b1/1+b2/3+b3/5+…+bn/2n次方-1,求数列bn的前n项和Tn
题目详情
2n次方=b1/1+b2/3+b3/5+…+bn/2n次方-1,求数列bn的前n项和Tn
▼优质解答
答案和解析
2n^2=b1/1+b2/3+b3/5+…+bn/(2n-1) (1)
n=1,b1=2
2(n-1)^2=b1/1+b2/3+b3/5+…+b(n-1)/(2n-3) (2)
(1)-(2)
bn/(2n-1) = 2(2n-1)
bn = 2(2n-1)^2
Tn =b1+b2+...+bn
consider
n^2 = n(n+1) -n
=(1/3)[n(n+1)(n+2)-(n-1)n(n+1)] -(1/2)[ n(n+1) -(n-1)n]
1^2+2^2+...+n^2
=(1/3)n(n+1)(n+2)-(1/2)n(n+1)
=(1/6)n(n+1)(2n+1)
bn = 2(2n-1)^2
Tn =b1+b2+...+bn
=2 .(1/6)(2n-1)[(2n-1)+1][2(2n-1)+1]
=(1/3)(2n-1)(2n)(4n-1)
=(2/3)n(2n-1)(4n-1)
n=1,b1=2
2(n-1)^2=b1/1+b2/3+b3/5+…+b(n-1)/(2n-3) (2)
(1)-(2)
bn/(2n-1) = 2(2n-1)
bn = 2(2n-1)^2
Tn =b1+b2+...+bn
consider
n^2 = n(n+1) -n
=(1/3)[n(n+1)(n+2)-(n-1)n(n+1)] -(1/2)[ n(n+1) -(n-1)n]
1^2+2^2+...+n^2
=(1/3)n(n+1)(n+2)-(1/2)n(n+1)
=(1/6)n(n+1)(2n+1)
bn = 2(2n-1)^2
Tn =b1+b2+...+bn
=2 .(1/6)(2n-1)[(2n-1)+1][2(2n-1)+1]
=(1/3)(2n-1)(2n)(4n-1)
=(2/3)n(2n-1)(4n-1)
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