數學考試題目

1/1+1/[1+(1+2)]+1/[1+(1+2)+(1+2+3)]+............+1/[1+(1+2)+.....(1+2+3+...+29+30)]
請問這題可以用 Sigma嗎
那要先做怎樣初步的處理??
Sol
先考慮一般項
1/[1+(1+2)+(1+2+3)+…+(1+2+3+…+k)]
=1/Σ(m=1 tok)_(1+2+,,,+m)
=2/Σ(m=1 tok)_[m(m+1)]
=2/Σ(m=1 tok)_(m^2+m)
=2/[k(k+1)(2k+1)/6+k(k+1)/2]
=2/[k(k+1)/6]*[2k+1+3]
=6/[k(k+1)(k+2)]
=a/k+b/(k+1)+c/(k+2)
6=a(k+1)(k+2)+bk(k+2)+ck(k+1)
when k=0
6=a(1)(2)
a=3
when k=－1
6=b(－1)(1)
b=－6
when k=－2
6=c(－2)(－1)
c=3
1/[1+(1+2)+(1+2+3)+…+(1+2+3+…+k)]
=3/k－6/(k+1)+3/(k+2)
So
1/1+1/[1+(1+2)]+1/[1+(1+2)+(1+2+3)]+............+1/[1+(1+2)+.....(1+2+3+...+29+30)]
=Σ(k=1 to 30)_ Σ1/[1+(1+2)+(1+2+3)+…+(1+2+3+…+k)]
=Σ(k=1 to 30)_ [3/k-6/(k+1)+3/(k+2)]
=(3/1+3/2+3/3+3/4+……+3/27+3/28+3/29+3/30)
－(6/2+6/3+6/4+6/5+…….+6/28+6/29+6/30+6/31)
+(3/3+3/4+3/5+3/6+…..+3/29+3/30+3/31+3/32)
=(3/1)－(3/31)－(3/2)+(3/32)
=1485/992

先考慮一般的情形

1+(1+2)+(1+2+3)+....(1+2+3+...+n)=Σ k(k+1)/2, k=1~n
=(1/2)Σ(k^2+k)
=(1/2)[n(n+1)(2n+1)/6+n(n+1)/2]
=n(n+1)(n+2)/6

原式=Σ 6/[k(k+1)(k+2)], k=1~30
=6Σ[1/k[1/(k+1)-1/(k+2)]
=6Σ[1/k-1/(k+1)-(1/2)(1/k-1/(k+2)]
=6{[(1-1/2)+(1/2-1/3)+.......(1/30-1/31)]
-(1/2)[(1-1/3)+(1/2-1/4)+(1/3-1/5)+.....(1/30)-1/32)]

大概就是這樣

切塊
切成1/1
1/1+(1+2)
1/1+(1+2)+(1+2+3)
.
.
.
1/1+(1+2)+.....(1+2+3+...+29+30)
第n塊為sigma m=1~n {(n-m+1)m}
然後把每塊加起來
sigma n=1~30 {sigma m=1~n {(n-m+1)m}}
然後先作m(把n當常數)
再作n可得解

http://tw.myblog.yahoo.com/sincos-heart/article?mid=5042