Cho Sn=1.2+3.4+5.6+...+2n−1.2n . Tính S100 biết rằng ∑i=1n2i=2+4+6+...2n=nn+1; ∑i=1ni2=1+22+32+...+n2=nn+12n+16

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

Cho Sn=1.2+3.4+5.6+...+(2n-1).2n. Tính S100 biết rằng