Cn+2(2)=45
(n+2)(n+1)/2=45
n²+3n+2=90
n²+3n-88=0
(n+11)(n-8)=0
n1=-11(不合題意,舍去)
n2=8
∴n=8Cn+2(2)=45(n+2)(n+1)/2=45這個(gè)怎么來(lái)的你是初中還是高中？初中那好吧，你這樣理單段線段有n+1條雙段線段有n條三段線段有n-1條(n+1)段線段有1條[(n+1)+1]*(n+1)/2=(n+2)(n+1)/2=45(首項(xiàng)加末項(xiàng)乘項(xiàng)數(shù)除以二)[(n+1)+1]*(n+1)/2）這個(gè)不太懂..單段線段有n+1條雙段線段有n條三段線段有n-1條(n+1)段線段有1條線段總數(shù)=n+1+n+n-1+……+1等差數(shù)列求和公式：(首項(xiàng)+末項(xiàng))*項(xiàng)數(shù)/2代入：[(n+1)+1]*(n+1)/2n+1是首項(xiàng)，1是末項(xiàng)，第二個(gè)n+1是項(xiàng)數(shù)，再除以2即為線段總數(shù)=45第二個(gè)n+1是項(xiàng)數(shù)?哪是第二個(gè)?[ (n+1)首項(xiàng) + 1末項(xiàng) ]* (n+1)項(xiàng)數(shù)/2