解22,42,62,.402
是22为首项,20为公差的等差数列,
即an=22+(n-1)×20
令an=402
即22+(n-1)×20=402
解得n=20
故22,42,62,.402有20项
故
22+42+62+...402
=20(22+402)/2
=4240.
解22,42,62,.402
是22为首项,20为公差的等差数列,
即an=22+(n-1)×20
令an=402
即22+(n-1)×20=402
解得n=20
故22,42,62,.402有20项
故
22+42+62+...402
=20(22+402)/2
=4240.