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the triangular up to 15stage, can you set triangular number​

Sagot :

Answer:

1 (1)

3 (1 + 2)

6 (1 + 2 + 3)

10 (1 + 2 + 3 + 4)

15 (1 + 2 + 3 + 4 + 5)

21 (1 + 2 + 3 + 4 + 5 + 6)

28 (1 + 2 + 3 + 4 + 5 + 6 + 7)

36 (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8)

45 (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)

55 (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10)

66 (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11)

78 (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12)

91 (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13)

105 (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14)

120 (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15)

Answer:

Here are the first 15 triangular numbers:

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120

Step-by-step explanation:

Triangular numbers are the sum of the first

n natural numbers. The formula for the

n-th triangular number is:

=

(

+

1

)

2

T

n

=

2

n(n+1)

Here’s the step-by-step calculation for the first 15 triangular numbers:

For

=

1

n=1:

1

=

1

(

1

+

1

)

2

=

1

×

2

2

=

1

T

1

=

2

1(1+1)

=

2

1×2

=1

For

=

2

n=2:

2

=

2

(

2

+

1

)

2

=

2

×

3

2

=

3

T

2

=

2

2(2+1)

=

2

2×3

=3

For

=

3

n=3:

3

=

3

(

3

+

1

)

2

=

3

×

4

2

=

6

T

3

=

2

3(3+1)

=

2

3×4

=6

For

=

4

n=4:

4

=

4

(

4

+

1

)

2

=

4

×

5

2

=

10

T

4

=

2

4(4+1)

=

2

4×5

=10

For

=

5

n=5:

5

=

5

(

5

+

1

)

2

=

5

×

6

2

=

15

T

5

=

2

5(5+1)

=

2

5×6

=15

For

=

6

n=6:

6

=

6

(

6

+

1

)

2

=

6

×

7

2

=

21

T

6

=

2

6(6+1)

=

2

6×7

=21

For

=

7

n=7:

7

=

7

(

7

+

1

)

2

=

7

×

8

2

=

28

T

7

=

2

7(7+1)

=

2

7×8

=28

For

=

8

n=8:

8

=

8

(

8

+

1

)

2

=

8

×

9

2

=

36

T

8

=

2

8(8+1)

=

2

8×9

=36

For

=

9

n=9:

9

=

9

(

9

+

1

)

2

=

9

×

10

2

=

45

T

9

=

2

9(9+1)

=

2

9×10

=45

For

=

10

n=10:

10

=

10

(

10

+

1

)

2

=

10

×

11

2

=

55

T

10

=

2

10(10+1)

=

2

10×11

=55

For

=

11

n=11:

11

=

11

(

11

+

1

)

2

=

11

×

12

2

=

66

T

11

=

2

11(11+1)

=

2

11×12

=66

For

=

12

n=12:

12

=

12

(

12

+

1

)

2

=

12

×

13

2

=

78

T

12

=

2

12(12+1)

=

2

12×13

=78

For

=

13

n=13:

13

=

13

(

13

+

1

)

2

=

13

×

14

2

=

91

T

13

=

2

13(13+1)

=

2

13×14

=91

For

=

14

n=14:

14

=

14

(

14

+

1

)

2

=

14

×

15

2

=

105

T

14

=

2

14(14+1)

=

2

14×15

=105

For

=

15

n=15:

15

=

15

(

15

+

1

)

2

=

15

×

16

2

=

120

T

15

=

2

15(15+1)

=

2

15×16

=120

So, the first 15 triangular numbers are:

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120