<Question>
Compute $99^2+99+1$ in your head.
</Question>

<think>
<Plan_1>
Okay, let's see. I need to compute 99 squared plus 99 plus 1 in my head. Hmm, how to approach this. Maybe breaking it down step by step.

<prompt_1.1> Break down the computation of 99 squared plus 99 plus 1 using the above plan. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
First, 99 squared. I remember that squaring numbers close to 100 can be done using some tricks. Like, (100 - 1)^2. Let me expand that. 100 squared is 10,000, then minus 2 times 100 times 1 which is 200, and then plus 1. So 10,000 - 200 + 1 = 9801. So 99 squared is 9801.

Then, adding 99 to that. 9801 + 99. Hmm, 9801 + 100 would be 9901, but since it's 99, that's one less. So 9901 - 1 = 9900. So now we have 9801 + 99 = 9900.

Then add 1 more. 9900 + 1 = 9901.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Wait, so is the answer 9901?

But let me check another way to make sure I didn't make a mistake. Maybe using the formula for n^2 + n + 1. That's similar to the formula for the sum of the first n natural numbers plus n^2, but not exactly. Wait, maybe there's a formula here.

Alternatively, maybe recognize that 99^2 + 99 + 1 can be written as 99(99 + 1) + 1.

<prompt_2.1> Verify the result using the formula 99(99 + 1) + 1 as per the above plan. </prompt_2.1>

Alternatively, maybe think of 99 as (100 - 1). Let's see:

<prompt_2.2> Verify the result using the binomial expansion of (100 - 1)^2 as per the above plan. </prompt_2.2>

Another way: Maybe add 99^2 + 99 first. Let's see, 99^2 is 9801, as before.

<prompt_2.3> Verify the result by directly adding 99^2 and 99, then adding 1 as per the above plan. </prompt_2.3>

But just to be thorough, let me compute it directly.

<prompt_2.4> Compute 99^2 + 99 + 1 directly as per the above plan. </prompt_2.4>

Alternatively, if I compute 99*100, which is 9900, then 99*100 +1 is 9900 +1=9901. Wait, that's a different way.

<prompt_2.5> Verify the result using the shortcut for 99*100 as per the above plan. </prompt_2.5>

Is there a pattern here? Maybe if I consider n=99, then n^2 +n +1 = (n+1)^2 -n.

<prompt_2.6> Identify any patterns or alternative methods for computing n^2 + n + 1 as per the above plan. </prompt_2.6>
</Plan_2>

<Execution_2>
<execution_2.1>
Let's try that. 99 times (99 + 1) is 99*100. Which is 9900. Then add 1, so 9900 + 1 = 9901.
</execution_2.1>

<execution_2.2>
99^2 + 99 + 1 = 99*99 + 99 + 1. Factor 99 out of the first two terms: 99*(99 + 1) + 1 = 99*100 + 1 = 9900 + 1 = 9901.
</execution_2.2>

<execution_2.3>
Then 9801 + 99. As before, that's 9900. Then 9900 +1 is 9901. Same result.
</execution_2.3>

<execution_2.4>
99 times 99. Let me compute 100*99 = 9900, then subtract 1*99, which is 99. So 9900 - 99 = 9801. Then add 99: 9801 + 99. As before, that's 9900. Then add 1: 9901.
</execution_2.4>

<execution_2.5>
Wait, 99^2 is 99*99. Then 99*99 + 99 is 99*(99 +1) = 99*100 = 9900. Then adding 1 gives 9901. Oh, that's a smarter way. So 99^2 +99 +1 = 99*(99 +1) +1 = 99*100 +1 = 9900 +1 =9901.
</execution_2.5>

<execution_2.6>
Let me check:

(n+1)^2 = n^2 + 2n +1. So if we subtract 3n from that, we get n^2 -n +1. Hmm, not quite. Wait, maybe another approach.

Alternatively, n^2 +n +1 = (n^2 - 2n +1) + 2n. Which is (n-1)^2 + 2n. Let's try that with n=99.

(n-1)^2 = 98^2. 98 squared is... 98*98. Let me compute that. 100*100=10,000. Subtract 2*100*2 + 2^2? Wait, maybe another way. 98 is 100 -2. So (100 -2)^2 = 100^2 -4*100 +4 = 10,000 -400 +4= 9604. So 98^2=9604.

Then, (n-1)^2 +2n = 9604 + 2*99 = 9604 + 198. Let's compute that. 9604 + 200 is 9804, subtract 2 gives 9802. Wait, but that contradicts the previous result. Hmm, so maybe that formula isn't correct. Wait, let me check.

Wait, the original expression is n^2 +n +1. If I write it as (n-1)^2 + 2n, then (n-1)^2 +2n = n^2 -2n +1 +2n = n^2 +1. Which is not equal to n^2 +n +1. So that approach was wrong. So that method was incorrect. So scratch that.

So going back, the first method seems solid. 99 squared is 9801, add 99 gives 9900, add 1 gives 9901. That seems correct.

Alternatively, let me compute 99*99 +99 +1. Factor 99 out of the first two terms: 99*(99 +1) +1 =99*100 +1=9900 +1=9901. Oh, that's a smarter way. So 99^2 +99 +1 =99*(99 +1) +1=99*100 +1=9900 +1=9901.

Yes, that's a more straightforward way. So by factoring 99 from the first two terms, it simplifies the calculation. That's a good method.

Another way to verify: Let's compute 99^2 +99 +1 step by step using another method. Let's compute 99^2 as (100 -1)^2 =100^2 -2*100*1 +1^2=10,000 -200 +1=9801. Then 9801 +99=9900. Then 9900 +1=9901. Yep, same result.

Alternatively, using decimal arithmetic. Let's compute 99*99:



So 9801 +99=9900, then +1=9901. Correct.

Another angle: Maybe using the formula for the sum of cubes or something, but I don't think that's applicable here. Wait, n^3 -1 factors into (n-1)(n^2 +n +1). So if we rearrange, n^2 +n +1 = (n^3 -1)/(n -1). But in this case, n=100? Wait, not sure.
</execution_2.6>
</Execution_2>

<Final_answer>
99^2 + 99 + 1 = 99 \times (99 + 1) + 1 = 99 \times 100 + 1 = 9900 + 1 = \boxed{9901}.

\]
</Final_answer>
</think>