The first term of a sequence is 2021. Each succeeding term is the sum of the squares of the digits of the previous term. What is the 2021th term in the sequence?
Answers
Given : The first term of a sequence is 2021. Each succeeding term is the sum of the squares of the digits of the previous term.
To Find : What is the 2021th term in the sequence
Solution:
1st term = 2021
2nd term = 2² + 0² + 2² + 1 = 4 + 0 + 4 + 1 = 9
3rd term = 9² = 81
4th term = 8² + 1² = 64 + 1 = 65
5th term = 6² + 5² = 36 + 25 = 61
6th term = 6² + 1² = 36 + 1 = 37
7th term = 3² + 7² = 9 + 49 = 58
8th term = 5² + 8² = 25 + 64 = 89
9th term = 8² + 9² = 64 + 81 = 145
10th term = 1² + 4² + 5² = 1 + 16 + 25 = 42
11th term = 4² + 2² = 16 + 4 = 20
12th term = 2² + 0² = 4
13th term = 4² = 16
14th term = 1² + 6² = 37
Sequence will repeat now
6th , 14th , 22 nd would be same
6th , 14th , 22nd -- 37 8k + 6th term (k is positive integer)
7th , 17th , 23rd -- 58 8k + 7th term
8th , 16th , 24th - 89 8k th term
9th , 17th , 25th 145 8k +1 term
10th , 18th , 26th 42 8k + 2 term
11th , 19th , 27th 20 8k + 3 term
12th , 20th , 28th 4 8k + 4 term
13th , 21st , 29th = 16 8k + 5th term
2021 = 252 * 8 + 5 = 8k + 5th term
Hence, 2021th term in the sequence is 16