Q.If $t_n$ denotes the $n$th term of the series $2 + 3 + 6 + 11 + 18 + \ldots$ then $t_{50}$ is
(A) $49^2 - 1$
(B) $49^2$
(C) $50^2 + 1$
(D) $49^2 + 2$

Concept and Intuition

We have a nested radical:

$a_n = \sqrt{7 + \sqrt{7 + \sqrt{7 + \cdots}}}$ with $n$ square roots.

This is a recursive sequence: each new term adds one more layer.

The key insight: such sequences either blow up or converge to a fixed point.

Here, because we keep adding a constant under a square root, the growth slows down.

We can find the limit by solving $L = \sqrt{7 + L}$, which gives $L^2 - L - 7 = 0$, so $L = \frac{1 + \sqrt{29}}{2} \approx 3.37$.

Since $a_1 = \sqrt{7} \approx 2.65$ and the sequence increases toward $3.37$, every term is less than $4$.

That immediately points to option (D).

Step-by-step reasoning

1. Define the sequence recursively

   Let $a_1 = \sqrt{7}$.

   For $n \geq 1$, $a_{n+1} = \sqrt{7 + a_n}$.

   This matches the given nested radical: $a_2 = \sqrt{7 + \sqrt{7}}$, etc.

2. Check monotonicity

   Compare $a_2$ and $a_1$:

   $a_2 = \sqrt{7 + \sqrt{7}} > \sqrt{7} = a_1$ because $7 + \sqrt{7} > 7$.

   Assume $a_k > a_{k-1}$. Then

   $a_{k+1} = \sqrt{7 + a_k} > \sqrt{7 + a_{k-1}} = a_k$.

   By induction, the sequence is strictly increasing.

3. Find an upper bound

   Solve $x = \sqrt{7 + x}$:

   $x^2 = 7 + x \implies x^2 - x - 7 = 0 \implies x = \frac{1 \pm \sqrt{29}}{2}$.

   The positive root is $L = \frac{1 + \sqrt{29}}{2} \approx 3.37$.

   Claim: $a_n < L$ for all $n$.

   Base: $a_1 = \sqrt{7} \approx 2.65 < L$.

   Inductive step: if $a_n < L$, then

   $a_{n+1} = \sqrt{7 + a_n} < \sqrt{7 + L} = L$ (since $L$ satisfies $L = \sqrt{7 + L}$).

   So $a_n < L$ for all $n$.

4. Conclude the range …

Concept and Intuition

If $\alpha,\beta$ are roots of $x^2-6x-2=0$, then $\alpha^2=6\alpha+2$ and $\beta^2=6\beta+2$. Multiplying by $\alpha^{n-2}$ and $\beta^{n-2}$ respectively and subtracting shows $a_n=\alpha^n-\beta^n$ obeys the recurrence $a_n=6a_{n-1}+2a_{n-2}$ — this avoids ever computing $\alpha,\beta$ explicitly.

Step-by-Step Solution

1. From $x^2-6x-2=0$: $\alpha^2=6\alpha+2,\ \beta^2=6\beta+2$.
2. Multiply by $\alpha^{n-2},\beta^{n-2}$: $\alpha^n=6\alpha^{n-1}+2\alpha^{n-2}$, $\beta^n=6\beta^{n-1}+2\beta^{n-2}$.
3. Subtract: $a_n=6a_{n-1}+2a_{n-2}$. …

Concept and Intuition

When $\alpha,\beta$ are roots of a quadratic $x^2=px+q$, any power sum/difference like $a_n=\alpha^n\pm\beta^n$ satisfies the same linear recurrence $a_n = p\,a_{n-1}+q\,a_{n-2}$, because each root individually obeys $\alpha^n = p\alpha^{n-1}+q\alpha^{n-2}$ (and similarly for $\beta$), and the recurrence is linear so it survives taking the difference.

Step-by-Step Solution

1. $\alpha,\beta$ are roots of $x^2-10x-8=0$, so $\alpha^2=10\alpha+8$ and $\beta^2=10\beta+8$; more generally $\alpha^n = 10\alpha^{n-1}+8\alpha^{n-2}$ and $\beta^n=10\beta^{n-1}+8\beta^{n-2}$ for $n\ge2$.
2. Subtracting: $\alpha^n-\beta^n = 10(\alpha^{n-1}-\beta^{n-1}) + 8(\alpha^{n-2}-\beta^{n-2})$, i.e. $a_n = 10a_{n-1}+8a_{n-2}$.
3. Apply with $n=10$: $a_{10} = 10a_9+8a_8$.
4. Rearranging: $a_{10}-8a_8 = 10a_9$. …