已知数列{an}为等差数列,d为公差,m,n,p,q∈N+,且m+n=p+q.求证:(1)am+an=ap+aq;(2)an=am+(n-m)d.
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已知数列{an}为等差数列,d为公差,m,n,p,q∈N+,且m+n=p+q.求证:(1)am+an=ap+aq;(2)an=am+(n-m)d.
已知数列{an}为等差数列,d为公差,m,n,p,q∈N+,且m+n=p+q.
求证:(1)am+an=ap+aq;(2)an=am+(n-m)d.
已知数列{an}为等差数列,d为公差,m,n,p,q∈N+,且m+n=p+q.求证:(1)am+an=ap+aq;(2)an=am+(n-m)d.
证明:由等差数列的性质,有
am=a1+(m-1)d.
an=a1+(n-1)d.
两式相加得
am+an=2a1+(m+n-2)d.①
ap=a1+(p-1)d.
aq=a1+(q-1)d.
两式相加得
ap+aq=2a1+(p+q-2)d.②
又因为m+n=p+q
所以m+n+2=p+q+2
因此①=②,即
am+an=ap+aq.
⑵an=a1+(n-1)d
am=a1+(m-1)d
两式相减得
an-am=(n-m)d.
移项得
an=am+(n-m)d.
原式得证!
(1)令a1为{an}的首项,则有
{an}的通项公式 an = a1 + (n-1)d
am + an = a1 + (m-1)d + a1 + (n-1)d = 2a1 + (m+n) -2d
ap + aq = a1 + (p-1)d + a1 + (q-1)d = 2a1 + (p+q) -2d
由于m+n=p+q,所以 2a1 + (m+n) -2d = ...
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(1)令a1为{an}的首项,则有
{an}的通项公式 an = a1 + (n-1)d
am + an = a1 + (m-1)d + a1 + (n-1)d = 2a1 + (m+n) -2d
ap + aq = a1 + (p-1)d + a1 + (q-1)d = 2a1 + (p+q) -2d
由于m+n=p+q,所以 2a1 + (m+n) -2d = 2a1 + (p+q) -2d
即 am+an=ap+aq
(2)an - am = a1 + (n-1)d - a1 - (m-1)d = (n-m)d
所以 an = am + (n-m)d
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