1+3=(1+3)*2/2;1+3+5=(1+5)*3;1+3+5+7=(1+7)*4/2````,则1+3+5+7+9+```+(2n-1)等于?a n*nb (n+1)*(n+1)c (1+n)*n/2d (n+1)*n/2

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1+3=(1+3)*2/2;1+3+5=(1+5)*3;1+3+5+7=(1+7)*4/2````,则1+3+5+7+9+```+(2n-1)等于?an*nb(n+1)*(n+1)c(1+n)*n/2

1+3=(1+3)*2/2;1+3+5=(1+5)*3;1+3+5+7=(1+7)*4/2````,则1+3+5+7+9+```+(2n-1)等于?a n*nb (n+1)*(n+1)c (1+n)*n/2d (n+1)*n/2
1+3=(1+3)*2/2;1+3+5=(1+5)*3;1+3+5+7=(1+7)*4/2````,则1+3+5+7+9+```+(2n-1)等于?
a n*n
b (n+1)*(n+1)
c (1+n)*n/2
d (n+1)*n/2

1+3=(1+3)*2/2;1+3+5=(1+5)*3;1+3+5+7=(1+7)*4/2````,则1+3+5+7+9+```+(2n-1)等于?a n*nb (n+1)*(n+1)c (1+n)*n/2d (n+1)*n/2
前三个可以看出规律
和=(首项+末项)*项数/2
项数=(末项+1)/2
所以 1+3+5+7+9+```+(2n-1)
=(1+2n-1)*(2n/2)/2
=2n*n/2
=n*n
选a

a
1+3+5+7+9+```+(2n-1)
=(1+2n-1)*(2n/2)/2
=2n*n/2
=n*n

a n*n