解:根据题意可知为首项为3,公差为2的等差数列所以前n项和为n x [3+(2n+1)]/2=n x (2n+4)/2=2n(n+2)/2=n(n+2)=n2+2n1+3+5+7+....+(2n-1)=【1+(2n-1)】x n /2=【1+2n-1】x n /2=2n x n /2=n x n=n^2扩展资料:等差数列的求和方法:方法是倒序相加Sn=1+2+3+……+(n-1)+nSn=n+(n-1)+(n-2)+……+2+1两式相加2Sn=(1+n)+(2+n-1)+(3+n-2)+……+(n-1+2)+(n+1)=(n+1)+(n+1)+(n+1)+……+(n+1)+(n+1)一共n项(n+1)2Sn=n(n+1)Sn=n(n+1)/2倒序相加是数列求和中一种常规方法1+3+5+...+(2n-1)=n^2最简单的方法是将己知式左边当成首项为1,公差为2的等差数列,代入等差数列求和公式即可。笨方法是:1+3+5+(2n-1)=(2×1-1)+(2×2-1)+(3×2-1)+...+(2n-1)=2×(1+2+3+...n)-1×n=2×[n(n+1)/2]-n=n^2+n-n=n^2。设S=1 +3 +5 +......+(2n-1)则S=(2n-1)+(2n-2)+(2n-3)+......+1对应项相加:2S=2n+2n+...+2n (n项)=2n*n=2n^2所以S=n^2约为4.43813 准确解太长,没法写