Question

求解第二问，详细的步骤，这种题目不是很懂，请求教

Answer 1

a(2) = 3a(1) = 3.

先分奇偶项分别求通项公式。
a(2k+2) = 3a(2k+1) = 3[a(2k) + 3^(k+1)] = 3a(2k) + 3^(k+2),
a(2k+2)/3^(k+1) = a(2k)/3^k + 3,
{a(2k)/3^k}是首项为a(2)/3 = 1, 公差为3的等差数列。
a(2k)/3^k = 1 + 3(k-1) = 3k-2.
a(2k) = (3k-2)3^k.

t(k)= a(2)+a(4)+...+a(2k) = (3*1-2)3 + (3*2-2)3^2 + (3*3-2)3^3 + ... + [3(k-1)-2]3^(k-1) + [3k-2]3^k,
3t(k) = (3*1-2)3^2 + (3*2-2)3^3 + ... + [3(k-1)-2]3^k + [3k-2]3^(k+1),

2t(k) = 3t(k) - t(k) = -(3*1-2)3 - 3*3^2 - 3*3^3 - ... - 3*3^k + [3k-2]3^(k+1)
= (3k-2)3^(k+1) + 6 - 3^2[1 + 3 + 3^2 + ... + 3^(k-1)]
= (3k-2)3^(k+1) + 6 - 9[3^k - 1]/(3-1)
= (3k-2)3^(k+1) + 6 - (3/2)3^(k+1) + 9/2
= [(6k -7)/2]3^(k+1) + 21/2

t(k) = [(6k-7)/4]3^(k+1) + 21/4 = [(6k-7)3^(k+1) + 21]/4

a(2k-1) = a(2k)/3 = (3k-2)3^(k-1).
a(3) = (3*2-2)3 = 12,
a(5) = (3*3-2)3^2 = 63.

a(1)+a(3)+...+a(2k-1) = t(k)/3 = [(6k-7)3^k + 7]/4.

s(2k) = a(2)+a(4)+...+a(2k) + a(1)+a(3)+...+a(2k-1) = [(6k-7)3^(k+1) + 21]/4 + [(6k-7)3^k + 7]/4
= (6k-7)3^k + 7.
s(2k-1) = s(2k) - a(2k) = (6k-7)3^k + 7 - (3k-2)3^k = (3k-5)3^k + 7

追问

谢谢

追答

楼主威武。。感谢采纳。。祝万事如意

Answer 2

金币不够啊

追答

通项公式求出，接下来自己弄吧，要分奇偶来