Evaluate the sum (1) ( 2 problems )

[shamieh]

Evaluate the sum
\(\displaystyle
\sum_{i=2}^{99}((i + 1)^2 - i^2))\)

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

\(\displaystyle 100^2 -2^2 = 9,996?\) Is this correct?

#2 Evaluate the sum

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

and got: \(\displaystyle 100^2 + 99^2 - 1 = 19,800\). Is this correct?

 

[MarkFL]

They are correct. Here are alternate approaches:

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

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

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

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