Cho Sn=1.2+3.4+5.6+...+(2n-1).2n. Tính S100 biết rằng
Giải thích
Ta có
Sn=∑i=1n2i2i−1=∑i=1n4i2−2i=4∑i=1ni2−∑i=1n2i =4n(n+1)(2n+1)6−n(n+1)=n(n+1)(4n−1)3⇒S100=100.(100+1)(4.100−1)3=1343300
Ta có
Sn=∑i=1n2i2i−1=∑i=1n4i2−2i=4∑i=1ni2−∑i=1n2i =4n(n+1)(2n+1)6−n(n+1)=n(n+1)(4n−1)3⇒S100=100.(100+1)(4.100−1)3=1343300