Question

求该数列的前n项和 an=n的平方除以2的n-1次方,求Sn

Answer 1

an =n^2 /2^(n-1)=2an -an
= n^2 /2^(n-2) - n^2/ 2^(n-1)
=(n-1)^2 /2^(n-2) - n^2/ 2^(n-1) + (2n-1)/2^(n-2)
设上式第三项为bn
bn= (2n-1)/2^(n-2) = 2bn -bn = (2n-1)/2^(n-3) - (2n-1)/2^(n-2)
== (2n-3)/2^(n-3) - (2n-1)/2^(n-2) + 2/2^(n-3)= (4n-6)/2^(n-2) - (4n-2)/2^(n-1) + 16/2^n
设 cn = (n^2+4n-2) / 2^(n-1)
则 an = c'n-1' - cn + 16/2^n
a'n-1' = c'n-2' - c'n-1' + 16/ 2^(n-1)
...
a1 = c0 -c1 + 16/2
上面n个式子加起来,注意右边第三项为等比数列
Sn = c0 - cn + 8 [1-(1/2)^n] / (1-1/2)
= -4 - (n^2+4n-2) / 2^(n-1) + 16(1- 1/2^n)
= 12 - 2 (n^2+4n+6)/2^n