Σの計算（基本） - オンライン講師ブログ

(1)

1. 次の数列の和を、𝛴 を使わず各項を書き並べて表せ。

1

1 2k2k++11 (

( )) 𝛴𝛴ⁿⁿ

k=1

k=1(( )) ((22)) 𝛴𝛴ⁿⁿ 22

k=1 k=1

k k

3

3 33 (

( )) 𝛴𝛴ⁿⁿ

k=1

k=1 ((44)) 𝛴𝛴ⁿⁿ kk

k=1 k=1

2 2

1

1 22--3k3k (

( )) 𝛴𝛴ⁿⁿ

k=1

k=1(( )) ((22)) 𝛴𝛴ⁿⁿ 33⋅⋅ --22

k=1

k=1 (( ))^k^k 3

3 bb++kk (

( )) 𝛴𝛴ⁿⁿ

k=1

k=1(( )) ((44)) 𝛴𝛴ⁿⁿ kk

k=1 k=1

3 3

3. 次の数列の初項から第 n 項までの和を 𝛴 を⽤いて表せ。

1

1 2 2,, 2 2,, 2 2,, 2 2,, ⋯⋯ (

( )) ((22)) 2 2,, 5 5,, 8 8,, 11 11,, ⋯⋯ 3

3 3 3,, 6 6,, 12 12,, 24 24,, ⋯⋯ (

( )) ((44)) 4 4,, 9 9,, 16 16,, 25 25,, ⋯⋯

1

1 2a 2a,, 2a 2a,, 2a 2a,, 2a 2a,, ⋯⋯ (

( )) ((22)) 30 30,, 26 26,, 22 22,, 18 18,, ⋯⋯ 3

3 ,, 1 1,, ,, ,, ⋯⋯ (

( )) 22 11 2 2

1 1 2

2 ((44)) 8 8,, 27 27,, 64 64,, 125 125,, ⋯⋯

5. 次の和を求めよ。

1

1 44 (

( )) 𝛴𝛴ⁿⁿ

k=1

k=1 ((22)) 𝛴𝛴ⁿⁿ 2k2k

k=1 k=1

3

3 kk++33 (

( )) 𝛴𝛴ⁿⁿ

k=1

k=1(( )) ((44)) 𝛴𝛴ⁿⁿ kk kk--11

k=1

k=1 (( )) 5

5 kk((kk --11 (

( )) 𝛴𝛴ⁿⁿ

k=1 k=1

2

2 )) ((66)) 𝛴𝛴ⁿⁿ 33

k=1 k=1

k-1 k-1

(2)

1

1 11 (

( )) 𝛴𝛴ⁿⁿ

k=1

k=1 ((22)) 𝛴𝛴ⁿⁿ 6k6k

k=1 k=1

3

3 11--kk (

( )) 𝛴𝛴ⁿⁿ

k=1

k=1(( )) ((44)) 𝛴𝛴ⁿⁿ ((kk ++kk--22

k=1 k=1

2

2 ))

5

5 44 kk--11 ((kk ++kk++11 (

( )) 𝛴𝛴ⁿⁿ

k=1

k=1 (( )) ²² )) ((66)) 𝛴𝛴ⁿⁿ 22

k=1 k=1

1-k 1-k

1

1 kk (

( )) ¹⁰¹⁰𝛴𝛴

k=1

k=1 ((22)) ²⁰²⁰𝛴𝛴ll

l=1 l=1

2 2

3 3 aa (

( )) ³⁰³⁰𝛴𝛴

a=1 a=1

3

3 ((44)) ¹⁰¹⁰𝛴𝛴 22

m=1 m=1

m m

1

1 3k3k--55 (

( )) ²⁰²⁰𝛴𝛴

k=1

k=1(( )) ((22)) ³⁰³⁰𝛴𝛴 ll++11 ll--11

l=1

l=1(( ))(( )) 3

3 aa++22 ((aa --2a2a++44 (

( )) ¹⁰¹⁰𝛴𝛴

a=1

a=1(( )) ²² )) ((44)) ¹⁰¹⁰𝛴𝛴

m=1 m=1

2 2

3 3

m-3 m-3

1

1 kk (

( )) ¹⁰¹⁰𝛴𝛴

k=5

k=5 ((22)) ²⁰²⁰𝛴𝛴 ll

k=10 k=10

2 2

3

3 2k2k--33 (

( )) 𝛴𝛴¹¹

k=1

k=1(( )) ((44)) 𝛴𝛴^m^m kk == mm mm++11

k=1 k=1

1 1 2

2 (( ))

(3)

1

1 2k2k++11 == 33++55++77++ ⋯⋯ ++ 2n2n++11 (

( )) 𝛴𝛴ⁿⁿ

k=1

k=1(( )) (( )) ((22)) 𝛴𝛴ⁿⁿ 22 ==22++22 ++22 ++ ⋯⋯ ++22

k=1 k=1

k

k 22 33 nn

3

3 33 ==33++33++33++ ⋯⋯ ++33 (

( )) 𝛴𝛴ⁿⁿ

k=1

k=1 ((44)) 𝛴𝛴ⁿⁿ kk ==11++22 ++33 ++ ⋯⋯ ++nn

k=1 k=1

2

2 22 22 22

1

1 22--3k3k (

( )) 𝛴𝛴ⁿⁿ

k=1

k=1(( ))

==((--11))++((--44))++((--77))++ ⋯⋯ ++((22--3n3n))

2

2 33⋅⋅ --22 (

( )) 𝛴𝛴ⁿⁿ

k=1

k=1 (( ))^k^k

==((--66))++1212++((--2424))++ ⋯⋯ ++33⋅⋅((--22))ⁿⁿ 3

3 bb++kk (

( )) 𝛴𝛴ⁿⁿ

k=1 k=1(( ))

== ((bb++11))++((bb++22))++((bb++33))++ ⋯⋯ ++((bb++nn))

4

4 kk ==11++88++2727++ ⋯⋯ ++nn (

( )) 𝛴𝛴ⁿⁿ

k=1 k=1

3

3 33

1

1 22 (

( )) 𝛴𝛴ⁿⁿ

k=1

k=1 ((22)) 𝛴𝛴ⁿⁿ 3k3k--22

k=1

k=1(( )) 3

3 33⋅⋅22 (

( )) 𝛴𝛴ⁿⁿ

k=1 k=1

n-1

n-1 ((44)) 𝛴𝛴ⁿⁿ kk++11

k=1

k=1(( ))³³

1

1 2a2a (

( )) 𝛴𝛴ⁿⁿ

k=1

k=1 ((22)) 𝛴𝛴ⁿⁿ --4k4k++3434

k=1

k=1(( )) 3

3 ( ( )) 𝛴𝛴ⁿⁿ

k=1 k=1

1 1 2 2

k-2 k-2

4

4 kk++11 (

( )) 𝛴𝛴ⁿⁿ

k=1

k=1(( ))²²

(4)

1

1 44 ==4n4n (

( )) 𝛴𝛴ⁿⁿ

k=1

k=1 ((22)) 𝛴𝛴ⁿⁿ 2k2k== nn nn++11

k=1

k=1 (( ))

3

3 kk++33 == nn nn++77 (

( )) 𝛴𝛴ⁿⁿ

k=1

k=1(( )) 11 2

2 (( )) ((44)) 𝛴𝛴ⁿⁿ kk --kk == nn nn++11 nn--11

k=1 k=1

2

2 11

3

3 (( ))(( )) 5

5 4k4k((kk --11 ==nn nn++11 nn--11 nn++22 (

( )) 𝛴𝛴ⁿⁿ

k=1 k=1

2

2 )) (( ))(( ))(( )) ((66)) 𝛴𝛴ⁿⁿ 33 ==

k=1 k=1

k-1

k-1 33 --11 2 2

n n

1

1 11 ==nn (

( )) 𝛴𝛴ⁿⁿ

k=1

k=1 ((22)) 𝛴𝛴ⁿⁿ 6k6k== 3n3n nn++11

k=1

k=1 (( ))

3

3 11--kk == nn 11--nn (

( )) 𝛴𝛴ⁿⁿ

k=1

k=1(( )) 11 2

2 (( )) ((44)) 𝛴𝛴ⁿⁿ ((kk ++kk--22 == nn nn--11 nn++44

k=1 k=1

2

2 )) 11

3

3 (( ))(( )) 5

5 44 kk--11 ((kk ++kk++11 (

( )) 𝛴𝛴ⁿⁿ

k=1

k=1 (( )) ²² ))

==nn((nn--11))((nn²²++3n3n++44))

6

6 22 ==22 11-- == 2 2-- (

( )) 𝛴𝛴ⁿⁿ

k=1 k=1

1-k

1-k 11

2 2ⁿⁿ

1 1 2 2^n-1^n-1

1

1 kk ==5555 (

( )) ¹⁰¹⁰𝛴𝛴

k=1

k=1 ((22)) ²⁰²⁰𝛴𝛴ll == 28702870

l=1 l=1

2 2

3

3 aa == 4410044100 (

( )) ²⁰²⁰𝛴𝛴

a=1 a=1

3

3 ((44)) ¹⁰¹⁰𝛴𝛴 22 ==20462046

m=1 m=1

m m

1

1 3k3k--55 == 530530 (

( )) ²⁰²⁰𝛴𝛴

k=1

k=1(( )) ((22)) ¹⁰¹⁰𝛴𝛴 ll++11 ll--11 == 375375

l=1

l=1(( ))(( ))

(5)

1

1 kk ==4545 (

( )) ¹⁰¹⁰𝛴𝛴

k=5

k=5 ((22)) ²⁰²⁰𝛴𝛴 kk == 24852485

k=10 k=10

2 2

3

3 2k2k--33 == --11 (

( )) 𝛴𝛴¹¹

k=1

k=1(( )) ((44)) 𝛴𝛴^m^m kk == mm mm++11

k=1 k=1

1 1 2

2 (( )) 10. 次の和を求めよ。

1

1 nn== nn (

( )) 𝛴𝛴ⁿⁿ

k=1 k=1

2

2 ((22)) 𝛴𝛴ⁿⁿ nn++11 ==nn nn++11

k=1

k=1(( )) (( )) 3

3 nn++kk == nn 3n3n++11 (

( )) 𝛴𝛴ⁿⁿ

k=1

k=1(( )) 11 2

2 (( )) ((44)) 𝛴𝛴ⁿⁿ nknk== nn nn++11

k=1 k=1

1 1 2

2 ²²(( ))