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DIA [1.3K]
3 years ago
8

A sequence is constructed according to the following rule: its first term is 7, and each next term is one more than the sum of t

he digits of the square of its previous term. Which number is on 2018th place?
Mathematics
1 answer:
Iteru [2.4K]3 years ago
7 0

Answer:

5

Step-by-step explanation:

According to the described rule, we have

a_1=7\\ \\a_1^2=7^2=49\Rightarrow a_2=4+9+1=14\\ \\a_2^2=14^2=196\Rightarrow a_3=1+9+6+1=17\\ \\a_3^2=17^2=289\Rightarrow a_4=2+8+9+1=20\\ \\a_4^2=20^2=400\Rightarrow a_5=4+0+0+1=5\\ \\a_5^2=5^2=25\Rightarrow a_6=2+5+1=8\\ \\a_6^2=8^2=64\Rightarrow a_7=6+4+1=11\\ \\a_7^2=11^2=121\Rightarrow a_8=1+2+1+1=5\\ \\\text{and so on...}

We can see the pattern

a_5=a_8=a_{11}=a_{14}=...=5\\ \\a_6=a_9=a_{12}=a_{15}=...=8\\ \\a_7=a_{10}=a_{13}=a_{16}=...=11

In other words, for all k\ge 2

a_{3k-1}=5\\ \\a_{3k}=8\\ \\a_{3k+1}=11

Now,

a_{2018}=a_{3\cdot 673-1}=5

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