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...
问题描述:
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
答
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