数列an中an=32sn=63,①若数列an为公差为1的等差数列,求a1②若数列an首相...数列an中an=32sn=63,①若数列an为公差为1的等差数列,求a1②若数列an首相为a1=1,求数列an平方的前n项和Tn

来源:学生作业帮助网 编辑:六六作业网 时间:2024/11/17 06:55:56
数列an中an=32sn=63,①若数列an为公差为1的等差数列,求a1②若数列an首相...数列an中an=32sn=63,①若数列an为公差为1的等差数列,求a1②若数列an首相为a1=1,求数列

数列an中an=32sn=63,①若数列an为公差为1的等差数列,求a1②若数列an首相...数列an中an=32sn=63,①若数列an为公差为1的等差数列,求a1②若数列an首相为a1=1,求数列an平方的前n项和Tn
数列an中an=32sn=63,①若数列an为公差为1的等差数列,求a1②若数列an首相...
数列an中an=32sn=63,①若数列an为公差为1的等差数列,求a1②若数列an首相为a1=1,求数列an平方的前n项和Tn

数列an中an=32sn=63,①若数列an为公差为1的等差数列,求a1②若数列an首相...数列an中an=32sn=63,①若数列an为公差为1的等差数列,求a1②若数列an首相为a1=1,求数列an平方的前n项和Tn
sn=n(a1+an)/2
63=n(a1+32)/2①
sn=n*a1+n*(n-1)*d/2
63=n*a1+n*(n-1)/2②
解①②式得
a1=-28
n=31

1.数列只有两项 a1=31,a2=32
2.数列如下:a1=1,a2=2,a3=4,a4=8,a5=16,a6=32
则平方的前n项和为(a1)2+(a2)2+(a3)2+(a4)2+(a5)2+(a6)2=1365
Tn=1*2*4*8*16*32=32768