Mathematic induction

[匿名]

2009-11-08 3:28 am

Please help me~~ thank you very much>< Dont know how to do because L.H.S
has n in each term.... help me to do it once plz....
Prove 1xn+2x(n-1)+3x(n-2)+...+(n-1)x2+nx1=n(n+1)(n+2)/6 is true for all positive
integers n.

回答 (1)

用戶2053

2009-11-08 3:54 am

✔ 最佳答案

(1)n+2(n-1)+3(n-2)+...+(n-1)2+n(1) = n(n+1)(n+2)/6

Assume the statement is true for all integers n :

when n = 1 :

LHS = 1

RHS = 1(1+1)(1+2)/6 = 1

RHS = LHS

Assume that when n = k is true ,

(1)k+2(k-1)+3(k-2)+...+(k-1)(2)+k(1) = k(k+1)(k+2)/6

when n = k+1 :

LHS =

(1)(k+1) + 2k + 3(k-1)+...+k(2) + (k+1)(1)

= (1)k+2(k-1)+3(k-2)+...+k(2-1) +(k+1)(1-1) + (1+2+3+...+k+(k+1) )

= k(k+1)(k+2)/6 + (1+2+3+...+k+(k+1) )

= k(k+1)(k+2)/6 + (1+k+1)(k+1)/2

= k(k+1)(k+2)/6 + (k+1)(k+2)/2

= k(k+1)(k+2)/6 + 3(k+1)(k+2)/6

= (k+1)(k+2)(k+3) / 6

which has the same format .

By MI we are done.

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