作业帮1.首项是a1,公差是d的等差数列的通项公式an=a1 (n-1)d,前n项和的公式是Sn=na1 (n(n-1))当n=k+1时,有Sn=(k+1)a1+((k+1)k)/2*d=ka1+(k(k-1))/2*d+a1+(k+1-1)d 这步不懂 谁能教教我吗当n=k+1时,有Sn=(k+1)a1+((k+1)k)/2*d=ka1+(k(k
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1.首项是a1,公差是d的等差数列的通项公式an=a1 (n-1)d,前n项和的公式是Sn=na1 (n(n-1))当n=k+1时,有Sn=(k+1)a1+((k+1)k)/2*d=ka1+(k(k-1))/2*d+a1+(k+1-1)d 这步不懂 谁能教教我吗当n=k+1时,有Sn=(k+1)a1+((k+1)k)/2*d=ka1+(k(k
1.首项是a1,公差是d的等差数列的通项公式an=a1 (n-1)d,前n项和的公式是Sn=na1 (n(n-1))
当n=k+1时,有Sn=(k+1)a1+((k+1)k)/2*d=ka1+(k(k-1))/2*d+a1+(k+1-1)d 这步不懂 谁能教教我吗
当n=k+1时,有Sn=(k+1)a1+((k+1)k)/2*d=ka1+(k(k-1))/2*d+a1+(k+1-1)d 这步不懂 谁能教教我吗
1.首项是a1,公差是d的等差数列的通项公式an=a1 (n-1)d,前n项和的公式是Sn=na1 (n(n-1))当n=k+1时,有Sn=(k+1)a1+((k+1)k)/2*d=ka1+(k(k-1))/2*d+a1+(k+1-1)d 这步不懂 谁能教教我吗当n=k+1时,有Sn=(k+1)a1+((k+1)k)/2*d=ka1+(k(k
a(n) = a + (n-1)d.
用数学归纳法证明,s(n) = na + n(n-1)d/2.
n=1时,s(1) = a(1) = a = 1*a + 1*(1-1)d/2,满足题意.
设n=k时,有s(n) = na + n(n-1)d/2.
则有,s(k) = ka + k(k-1)d/2.
当n=k+1时,s(k+1) = s(k) + a(k+1) = ka + k(k-1)d/2 + [a + kd] = (k+1)a + kd[(k-1)/2 + 1]
= (k+1)a + kd(k+1)/2
= (k+1)a + (k+1)(k+1-1)d/2,
也满足题意.
因此,由归纳法知,总有,s(n) = na + n(n-1)d/2.
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楼主问题中,第一个等号说的是,当n=k+1时,要证明的结论；
第二个等号说的是,s(k+1) = s(k) + a(k+1).
俺说清楚点没.