次のΣ計算をしなさい。
(1) |
|
1 .
k(k+2) |
(2) |
|
k−1 .
k(k+1)(k+2) |
(3) |
|
k−1
.
k(k+1)(k+2)(k+3) |
|
[分数は部分分数に分解・・・なんでもかんでも恒等式はよくありません。それなりの変形で簡単になります]
(1) |
|
1 .
k(k+2) |
= |
|
1
2 |
( |
1
k |
− |
1 .
k+2 |
) |
= |
1
2 |
|
( |
1
k |
− |
1 .
k+2 |
) |
= |
1
2 |
{( |
1+ |
1
2 |
+ |
1
3 |
+・・・+ |
1
n |
) |
− |
( |
1
3 |
+ |
1
4 |
+・・・+ |
1
n |
+ |
1 .
n+1 |
+ |
1 .
n+2 |
)} |
= |
1
2 |
( |
1+ |
1
2 |
− |
1 .
n+1 |
− |
1 .
n+2 |
) |
[
ここからの計算に工夫] |
= |
1
2 |
{( |
1− |
1
n+1 |
) |
+ |
( |
1
2 |
− |
1 .
n+2 |
)} |
= |
1
2 |
( |
n .
n+1 |
+ |
n .
2(n+2) |
) |
= |
n(3n+5) .
4(n+1)(n+2) |
(2) |
|
k−1 .
k(k+1)(k+2) |
= |
|
( |
1 .
(k+1)(k+2) |
− |
1 .
k(k+1)(k+2) |
) |
= |
|
{( |
1 .
k+1 |
− |
1
k+2 |
)− |
1
2 |
( |
1 .
k(k+1) |
− |
1 .
(k+1)(k+2) |
)} |
= |
( |
1
2 |
− |
1 .
n+2 |
)− |
1
2 |
( |
1
2 |
− |
1 .
(n+1)(n+2) |
) |
= |
1
4 |
− |
1
n+2 |
+ |
1 .
2(n+1)(n+2) |
= |
1
4 |
− |
2n+2−1 .
2(n+1)(n+2) |
= |
(n+1)(n+2)−2(2n+1)
4(n+1)(n+2) |
(3) |
|
k−1
.
k(k+1)(k+2)(k+3) |
= |
|
( |
1 .
(k+1)(k+2)(k+3) |
− |
1
.
k(k+1)(k+2)(k+3) |
) |
= |
1
2 |
|
( |
1 .
(k+1)(k+2) |
− |
1 .
(k+2)(k+3) |
) |
− |
1
3 |
|
( |
1 .
k(k+1)(k+2) |
− |
1 .
(k+1)(k+2)(k+3) |
) |
= |
1
2 |
( |
1
6 |
− |
1 .
(n+2)(n+3) |
) |
− |
1
3 |
( |
1
6 |
− |
1 .
(n+1)(n+2)(n+3) |
) |
= |
1
2 |
・ |
(n+2)(n+3)−6
6(n+2)(n+3) |
) |
− |
1
3 |
・ |
(n+1)(n+2)(n+3)−6
6(n+1)(n+2)(n+3) |
) |
= |
1
2 |
・ |
n(n+5) .
6(n+2)(n+3) |
− |
1
3 |
・ |
|
= |
n
.
36(n+1)(n+2)(n+3) |
・ |
{3(n+1)(n+5)−2(n | 2
| +6n+11)} |
= |
n
(n−1)(n+7) .
36(n+1)(n+2)(n+3) |
1,2,3,・・・,n からなる
n個の数列について次のものを求めよ。
(1) これら
n個のなかから異なる2個を選んでその積をつくる。この積の総和を求めよ。
(2) これら
n個のなかから異なる3個を選んでその積をつくる。この積の総和を求めよ。(やや難) |
解答例 |
(1) 求める和Sは、{(1+2+3+・・・+n) |
2 | −(1 |
2 | +2 |
2 | +3 |
2 |
+・・・+n |
2 |
)}÷2 と表せる。 |
S= |
1
2 |
{ |
1
4 |
n |
2
| (n+1) |
2
| − |
1
6 |
n(n+1)(2n+1) |
} |
= |
n(n+1)
24 |
{ |
3n(n+1)−2(2n+1) |
} |
= |
n(n+1)(n−1)(3n+2)
24 |
|