Recursive Sequence Sum

The difference between the $n$ th perfect square and the $n-1$ th perfect square is always the $n$ th (positive) odd number. Note that the $2n-1$ in the equation represents that odd number and $a_{n-1}$ is the previous perfect square (starting from $a_0=0$ ). So, $a_n=n^2$ ; the question is just asking us for the sum of the first 100 perfect squares , which is $\frac{(100)(100+1)(2\cdot100+1)}{6}=\boxed{338350}$

The first few $a_n$ show that $a_n = n^2$ , let us prove it by induction if it is true for all $n > 0$ .

1. For $n = 1$ , then $a_1 = a_0 + 2(1) - 1 = 0 + 2 - 1 = 1 = 1^2\implies$ the claim is true for $n=1$ .
2. Assuming that the claim is true for any $n$ , then:

$\begin{aligned} \quad a_{n+1} & = a_n + 2(n+1) - 1 \\ & = n^2 + 2n + 1 \\ & = (n+1)^2 \end{aligned}$

$\quad$ The claim is also true for $n+1$ .

We have, by induction, the claim $a_n = n^2$ is true for all $n > 0$ . Therefore,

$\begin{aligned} \sum_{n=1}^{100} a_n & = \sum_{n=1}^{100} a^2 \\ & = \frac{100(101)(201}{6} \\ & = \boxed{338350} \end{aligned}$

Take summation on both sides of given recurrence relation .

Then $a_n - a_0 = n^{2}$ .

Therefore $a_n= n^{2}$ .

Now again taking summation on both sides from n=1 to 100,

We get $S_{req} =( 100\times 101 \times 201)/6$ .