Xét (1−1)2n+1=C02n+1−C12n+1+C22n+1−C32n+1+...−C2n+12n+1=0
⇒C02n+1+C22n+1+...+C2n2n+1=C12n+1+C32n+1+...+C2n+12n+1
⇒2[C12n+1+C32n+1+...+C2n+12n+1]=[C02n+1+C22n+1+...+C2n2n+1]+[C12n+1+C32n+1+...+C2n+12n+1]
=C02n+1+C12n+1+C22n+1+C32n+1+...+C2n+12n+1=(1+1)2n+1=22n+1
⇒C12n+1+C32n+1+...+C2n+12n+1=22n=1024=210
⇒n=5⇒n2=25