1/2,3/2×2,5/2×2×2,…,2n-1/2×2×2×…×2(9个2),求前n向和Sn表达式
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1/2,3/2×2,5/2×2×2,…,2n-1/2×2×2×…×2(9个2),求前n向和Sn表达式
2^n --2的n次幂
“2n-1/2×2×2×…×2(9个2)”应该是n个2吧?
an = (2n-1)/2^n = 2n/2^n - 1/2^n
Sn = a1+a2+a3+...+an = 2(1/2^1 + 2/2^2 + 3/2^3 + ...+ n/2^n)
-(1-1/2^n)
2(1/2^1 + 2/2^2 + 3/2^3 + ...+ n/2^n) --记为Tn
则0.5Tn = 2[1/2^2 + 2/2^3 + 3/2^4 +...+ n/2^(n+1)]
相减 ,0.5Tn = 2 *[1 - 1/2^n - n/2^(n+1)]
Tn = 4*[1 - 1/2^n - n/2^(n+1)]
Sn = Tn - (1-1/2^n) = 3 *(1-1/2^n)-4 *n/2^(n+1) = 3 - (3+2*n)/2^n
“2n-1/2×2×2×…×2(9个2)”应该是n个2吧?
an = (2n-1)/2^n = 2n/2^n - 1/2^n
Sn = a1+a2+a3+...+an = 2(1/2^1 + 2/2^2 + 3/2^3 + ...+ n/2^n)
-(1-1/2^n)
2(1/2^1 + 2/2^2 + 3/2^3 + ...+ n/2^n) --记为Tn
则0.5Tn = 2[1/2^2 + 2/2^3 + 3/2^4 +...+ n/2^(n+1)]
相减 ,0.5Tn = 2 *[1 - 1/2^n - n/2^(n+1)]
Tn = 4*[1 - 1/2^n - n/2^(n+1)]
Sn = Tn - (1-1/2^n) = 3 *(1-1/2^n)-4 *n/2^(n+1) = 3 - (3+2*n)/2^n
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