Question

I need help with #47 to #50 ASAP…. Please help me with it

Answer 1

Step-by-step explanation:

a triangular number n is the sum of all natural numbers <= n.

t1 = 1

t2 = 1+2 = 3

t3 = 1+2+3 = 6

t4 = 1+2+3+4 = 10

...

so,

tn = tn-1 + n

47.

1×8 + 1 = 9 is a square number.

3×8 + 1 = 25 is a square number

6×8 + 1 = 49 is a square number

10×8 + 1 = 81 is a square number

48.

1/3 = 0 remainder 1

3/3 = 1 remainder 0

6/3 = 2 remainder 0

10/3 = 3 remainder 1

15/3 = 5 remainder 0

21/3 = 7 remainder 0

28/3 = 9 remainder 1

so, there seems to be a pattern 1 0 0 1 0 0 1 0 0 1 ...

49.

1/4 = 0 remainder 1

4/4 = 1 remainder 0

9/4 = 2 remainder 1

16/4 = 4 remainder 0

25/4 = 6 remainder 1

36/4 = 9 remainder 0

49/4 = 12 remainder 1

so, there seems to be a pattern 1 0 1 0 1 0 1 0 1 0 1 ...

50.

polygonal numbers is the real name for this.

the formula for dimensions = 5 is

(3n² − n)/2

for dimensions = 6 it is

2n² - n

so, dimensions=5 (and therefore dividing also by 5) we get the remainders

1/5 = 0 remainder 1

5/5 = 1 remainder 0

12/5 = 2 remainder 2

22/5 = 4 remainder 2

35/5 = 7 remainder 0

51/5 = 10 remainder 1

70/5 = 14 remainder 0

92/5 = 18 remainder 2

117/5 = 23 remainder 2

145/5 = 29 remainder 0

here the pattern is 1 0 2 2 0 1 0 2 2 0 1 0 2 2 0 ...

dimensions=6 (and therefore dividing also by 6) we get the remainders

1/6 = 0 remainder 1

6/6 = 1 remainder 0

15/6 = 2 remainder 3

28/6 = 4 remainder 4

45/6 = 7 remainder 3

66/6 = 11 remainder 0

91/6 = 15 remainder 1

120/6 = 20 remainder 0

153/6 = 25 remainder 3

190/6 = 31 remainder 4

231/6 = 38 remainder 3

276/6 = 46 remainder 0

325/6 = 54 remainder 1

here the pattern is 1 0 3 4 3 0 1 0 3 4 3 0 1 0 3 4 3 0 ...