Question

Let $T_n$ be the $n^{th}$ term of a series for $n \ge 1$ and $T_1=1, T_{n+1}=2T_n+4^n, n \ge 1$. If $S_n=$ Sum of first $n$ terms of the series and $S_n=\frac{a.4^n+b.2^{n+1}}{3}$ where $a,b \in I$, then the value of $|\frac{b}{a}|$ is

Answer 1

3/2

Answer 2

The recurrence relation is given by $T_{n+1} = 2T_n + 4^n$ for $n \ge 1$, with $T_1 = 1$.

We can solve this linear first-order non-homogeneous recurrence relation. Divide the recurrence by $2^{n+1}$:

$\frac{T_{n+1}}{2^{n+1}} = \frac{2T_n}{2^{n+1}} + \frac{4^n}{2^{n+1}}$

$\frac{T_{n+1}}{2^{n+1}} = \frac{T_n}{2^n} + \frac{(2^2)^n}{2^{n+1}} = \frac{T_n}{2^n} + \frac{2^{2n}}{2^{n+1}} = \frac{T_n}{2^n} + 2^{n-1}$.

Let $U_n = \frac{T_n}{2^n}$. Then $U_{n+1} = U_n + 2^{n-1}$ for $n \ge 1$.

We have $U_1 = \frac{T_1}{2^1} = \frac{1}{2}$.

For $n \ge 2$, we can write $U_n$ as a sum:

$U_n = U_1 + \sum_{k=1}^{n-1} (U_{k+1} - U_k) = U_1 + \sum_{k=1}^{n-1} 2^{k-1}$.

The sum is a geometric series: $\sum_{k=1}^{n-1} 2^{k-1} = 2^0 + 2^1 + \dots + 2^{n-2} = \frac{1(2^{n-1} - 1)}{2 - 1} = 2^{n-1} - 1$.

So, for $n \ge 2$, $U_n = \frac{1}{2} + (2^{n-1} - 1) = 2^{n-1} - \frac{1}{2}$.

We can check if this formula holds for $n=1$: $U_1 = 2^{1-1} - \frac{1}{2} = 2^0 - \frac{1}{2} = 1 - \frac{1}{2} = \frac{1}{2}$. The formula holds for $n \ge 1$.

Now, we find $T_n$ using $U_n = \frac{T_n}{2^n}$, so $T_n = U_n \cdot 2^n = (2^{n-1} - \frac{1}{2}) \cdot 2^n = 2^{n-1} \cdot 2^n - \frac{1}{2} \cdot 2^n = 2^{2n-1} - 2^{n-1}$.

Next, we find the sum of the first $n$ terms, $S_n = \sum_{k=1}^n T_k$.

$S_n = \sum_{k=1}^n (2^{2k-1} - 2^{k-1}) = \sum_{k=1}^n 2^{2k-1} - \sum_{k=1}^n 2^{k-1}$.

The first sum is $\sum_{k=1}^n 2^{2k-1} = 2^1 + 2^3 + 2^5 + \dots + 2^{2n-1}$. This is a geometric series with first term $A_1 = 2^1 = 2$, common ratio $R_1 = 2^2 = 4$, and $n$ terms.

The sum is $\frac{A_1(R_1^n - 1)}{R_1 - 1} = \frac{2(4^n - 1)}{4 - 1} = \frac{2(4^n - 1)}{3} = \frac{2 \cdot 4^n - 2}{3}$.

The second sum is $\sum_{k=1}^n 2^{k-1} = 2^0 + 2^1 + 2^2 + \dots + 2^{n-1}$. This is a geometric series with first term $A_2 = 2^0 = 1$, common ratio $R_2 = 2$, and $n$ terms.

The sum is $\frac{A_2(R_2^n - 1)}{R_2 - 1} = \frac{1(2^n - 1)}{2 - 1} = 2^n - 1$.

Now, substitute these sums back into the expression for $S_n$:

$S_n = \frac{2 \cdot 4^n - 2}{3} - (2^n - 1) = \frac{2 \cdot 4^n - 2 - 3(2^n - 1)}{3} = \frac{2 \cdot 4^n - 2 - 3 \cdot 2^n + 3}{3} = \frac{2 \cdot 4^n - 3 \cdot 2^n + 1}{3}$.

We are given that $S_n = \frac{a \cdot 4^n + b \cdot 2^{n+1}}{3}$.

Comparing this with $S_n = \frac{2 \cdot 4^n - 3 \cdot 2^n + 1}{3}$, and assuming that we ignore the constant term '+1' in our derived formula, we could try to match the exponential terms.

$\frac{a \cdot 4^n + b \cdot 2^{n+1}}{3} = \frac{a \cdot 4^n + 2b \cdot 2^n}{3}$.

Comparing with $\frac{2 \cdot 4^n - 3 \cdot 2^n}{3}$ (ignoring the constant term), we would get $a=2$ and $2b=-3$, so $b=-3/2$. This still does not give integer values for $a$ and $b$.

The problem statement likely has a typo in the form of $S_n$. If we assume that the formula for $S_n$ should match the exponential terms of our derived sum, ignoring the constant term, then we compare $a \cdot 4^n + b \cdot 2^{n+1}$ with $2 \cdot 4^n - 3 \cdot 2^n$.

$a \cdot 4^n + 2b \cdot 2^n = 2 \cdot 4^n - 3 \cdot 2^n$.

This implies $a=2$ and $2b=-3$, so $b=-3/2$. Still not integers.

Let's assume there is a typo in the exponent of $2$ in the given formula, and it should be $2^n$.

If $S_n = \frac{a \cdot 4^n + b \cdot 2^n}{3}$, then comparing with $S_n = \frac{2 \cdot 4^n - 3 \cdot 2^n + 1}{3}$, we get $a=2, b=-3$. These are integers.

In this case, $|\frac{b}{a}| = |\frac{-3}{2}| = \frac{3}{2}$.