Question

please answer !!!!!!!​

Answer 1

$\Large\text{\underline{\underline{Step 1. Check the conditions}}}$

The formula for the last two digits is -

$\text{ \cdots\longrightarrow The remainder of 7^{n}\div100. }$

We would get, -

$\text{ \cdots\longrightarrow Q as the quotient and R as the remainder.}$

In expression, -

$\cdots\longrightarrow7^{n}=100Q+R.$

However, if we multiply 7,

$\cdots\longrightarrow7^{n+1}=700Q+7R.$

We only need to check the remainder.

$\Large\text{\underline{\underline{Step 2. Recurring sequence}}}$

The last two digits repeat -

$\text{ \cdots\longrightarrow\boxed{\begin{aligned}&a_{1}=7\\&a_{2}=49\\&a_{3}=43\\&a_{4}=1.\end{aligned}} }$

Now we should know that -

$\text{ \cdots\longrightarrow\boxed{\begin{aligned}&a_{4n-3}=7\\&a_{4n-2}=49\\&a_{4n-1}=43\\&a_{4n}=1.\end{aligned}} }$

$\Large\text{\underline{\underline{Step 3. Conclusion part}}}$

(Choice 1)

Hence, -

$\cdots\longrightarrow a_{2021}=a_{4\times505+1}=a_{1}=7.$

(Choice 2 or 4)

And, -

$\text{ \cdots\longrightarrow\boxed{\begin{aligned}\displaystyle&\sum^{100}_{r=1}a_{r}\\\\&=\displaystyle\sum^{25}_{r=1}a_{4r-3}+\sum^{25}_{r=1}a_{4r-2}+\sum^{25}_{r=1}a_{4r-1}+\sum^{25}_{r=1}a_{4r}\\\\&=25\times(7+49+43+1)\\\\&=25\times 100\\\\&=2500.\end{aligned}} }$

(Choice 3)

Lastly, -

$\cdots\longrightarrow a_{2022}=a_{4\times505+2}=a_{2}=49.$

So, the last choice is correct.