Find the sum of odd integers from 1 to 2001.A1001000B1000000C1002001D1003002Answer: C. 1002001 Read Explanation: 1,3,5,7,....,2001Sn=n2[t1+tn]S_n=\frac{n}{2}[t_1+t_n]Sn=2n[t1+tn]a+(n−1)d=tna+(n-1)d=t_na+(n−1)d=tn1+(n−1)2=20011+(n-1)2=20011+(n−1)2=2001(n−1)2=2000(n-1)2=2000(n−1)2=2000n-1=1000n=1001Sn=10012[1+2001]S_n=\frac{1001}{2}[1+2001]Sn=21001[1+2001]=10012×2002=\frac{1001}{2}\times2002=21001×2002=1001×1001=1001 \times1001=1001×1001=1002001=1002001=1002001 Read more in App