MHB Evaluate the sum (1) ( 2 problems )

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The discussion evaluates two sums involving squared terms. The first sum, from 2 to 99, simplifies to 9996, confirmed by multiple methods. The second sum, from 2 to 100, results in 19800, also verified through different approaches. Both calculations demonstrate the effectiveness of algebraic manipulation in simplifying series. The final results for the sums are 9996 and 19800, respectively.
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Evaluate the sum
$$
\sum_{i=2}^{99}((i + 1)^2 - i^2))$$

So I found the pattern and got
$$
((3^2) - 2^2)) + ((4)^2 - (3)^2)) + ((5^2) - (4)^2)$$ ... etc etc

$$100^2 -2^2 = 9,996?$$ Is this correct?

#2 Evaluate the sum

$$
\sum_{i=2}^{100}(i^2 -(i - 2)^2) $$

and got: $$100^2 + 99^2 - 1 = 19,800$$. Is this correct?
 
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They are correct. Here are alternate approaches:

1.) $$S=\sum_{k=2}^{99}\left((k+1)^2-k^2 \right)$$

$$S=\sum_{k=2}^{99}\left(2k+1 \right)=99(100)-2+98=9996$$

2.) $$S=\sum_{k=2}^{100}\left(k^2-(k-2)^2 \right)$$

$$S=4\sum_{k=1}^{99}\left(k \right)=2\cdot99(100)=19800$$
 
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