兩題關於級數求和的問題

Question

(1) 1*n+2(n-1)+3(n-2)+......+n*1=??


(2) 1+(1+(1+2))+(1+(1+2)+(1+2+3))+......+(1+(1+2)+(1+2+3)+......+(1+2+3+......+n)=??

Answer 1

n
Σ k * ( n + 1 - k ) = Σ nk + k - k^2 = Σnk + Σk - Σk^2
k=1
=> Σnk = n * Σk = n * [ ( n )( n + 1 ) / 2 ]
Σk = ( n )( n + 1 ) / 2
Σk^2 = ( n ) * ( n + 1 ) ( 2n + 1 ) / 6
=> n * [ ( n )( n + 1 ) / 2 ] + ( n )( n + 1 ) / 2 -(n)*( n + 1 ) ( 2n + 1 ) / 6
= [ ( n )( n + 1 ) / 2 ] * [ n + 1 - ( 2n + 1 ) / 3 ]
= [ ( n )( n + 1 ) / 2 ] * [ ( n + 2 ) / 3 ]
= ( n ) ( n + 1 ) ( n + 2 ) / 6
2. 先算出這東西的第 an 項
a1項是 1 . a2 項是 1 + ( 1 + 2 )
n
Σk = n ( n + 1 ) / 2
k=1
=>第 an項就是
n
Σ k ( k + 1 ) / 2 = Σ ( k^2 + Σk ) / 2 = n * ( n + 1 ) ( n + 2 ) / 6
k=1
所以這個級數的和就是
n
Σ k * ( k + 1 ) ( k + 2 ) / 6 = ( Σ k^3 + 3Σk^2 + 2Σk ) / 6
k=1
Σk^3 = [ n * ( n + 1 ) / 2 ]^2
Σk^2 = n ( n + 1 ) ( 2n + 1 ) / 6
Σk = n ( n + 1 ) / 2
=>( Σ k^3 + 3Σk^2 + 2Σk ) / 6
= { [ n * ( n + 1 ) / 2 ]^2 + n ( n + 1 ) ( 2n + 1 ) / 2 + ( n ) ( n + 1 ) }/ 6
= [ ( n ) ( n + 1 ) ( n + 2 ) ( n + 3 ) / 4 ] / 6
= ( n ) ( n + 1 ) ( n + 2 ) ( n + 3 ) / 24

Answer 2

這一題已經有人問過了,
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http//tw.knowledge.yahoo.com/question/question?qid=1007102806594