Question

If $S_n = \sum_{r=1}^{n}\sum_{t=0}^{r-1}(\frac{1}{6^n}{}^nC_r{}^rC_t4^t)$, then value of $\ell$ where $\ell = \frac{1}{2}\sum_{n=1}^{\infty}(1-S_n)$ is

Answer 1

5/2

Answer 2

To find the value of $\ell$, we first need to simplify the expression for $S_n$.

Given: $S_n = \sum_{r=1}^{n}\sum_{t=0}^{r-1}(\frac{1}{6^n}{}^nC_r{}^rC_t4^t)$

We can factor out $\frac{1}{6^n}$: $S_n = \frac{1}{6^n}\sum_{r=1}^{n}\sum_{t=0}^{r-1}{}^nC_r{}^rC_t4^t$

Let's evaluate the inner sum first: $\sum_{t=0}^{r-1}{}^rC_t4^t$. We know the binomial expansion formula: $\sum_{t=0}^{r}{}^rC_t x^t = (1+x)^r$. So, $\sum_{t=0}^{r-1}{}^rC_t4^t = \left(\sum_{t=0}^{r}{}^rC_t4^t\right) - {}^rC_r4^r$. $\sum_{t=0}^{r-1}{}^rC_t4^t = (1+4)^r - 1 \cdot 4^r = 5^r - 4^r$. This simplification is valid for $r \ge 1$. Since the outer sum for $r$ starts from $1$, this is applicable.

Now substitute this back into the expression for $S_n$: $S_n = \frac{1}{6^n}\sum_{r=1}^{n}{}^nC_r(5^r - 4^r)$ $S_n = \frac{1}{6^n}\left(\sum_{r=1}^{n}{}^nC_r5^r - \sum_{r=1}^{n}{}^nC_r4^r\right)$

Let's evaluate each sum separately using the binomial theorem $\sum_{k=0}^{n}{}^nC_k x^k = (1+x)^n$:

1. $\sum_{r=1}^{n}{}^nC_r5^r$: This sum is equal to $\left(\sum_{r=0}^{n}{}^nC_r5^r\right) - {}^nC_05^0$. $\sum_{r=1}^{n}{}^nC_r5^r = (1+5)^n - 1 \cdot 1 = 6^n - 1$.

2. $\sum_{r=1}^{n}{}^nC_r4^r$: This sum is equal to $\left(\sum_{r=0}^{n}{}^nC_r4^r\right) - {}^nC_04^0$. $\sum_{r=1}^{n}{}^nC_r4^r = (1+4)^n - 1 \cdot 1 = 5^n - 1$.

Substitute these back into the expression for $S_n$: $S_n = \frac{1}{6^n}\left((6^n - 1) - (5^n - 1)\right)$ $S_n = \frac{1}{6^n}(6^n - 1 - 5^n + 1)$ $S_n = \frac{1}{6^n}(6^n - 5^n)$ $S_n = 1 - \frac{5^n}{6^n} = 1 - \left(\frac{5}{6}\right)^n$.

Now we need to find the value of $\ell = \frac{1}{2}\sum_{n=1}^{\infty}(1-S_n)$. First, calculate $(1-S_n)$: $1 - S_n = 1 - \left(1 - \left(\frac{5}{6}\right)^n\right) = \left(\frac{5}{6}\right)^n$.

Now substitute this into the expression for $\ell$: $\ell = \frac{1}{2}\sum_{n=1}^{\infty}\left(\frac{5}{6}\right)^n$.

This is an infinite geometric series with the first term $a = \frac{5}{6}$ (for $n=1$) and common ratio $r = \frac{5}{6}$. The sum of an infinite geometric series $\sum_{n=1}^{\infty} ar^{n-1}$ (or $\sum_{n=1}^{\infty} ar^n$ with appropriate $a$) is $\frac{\text{first term}}{1-\text{common ratio}}$, provided $|r|<1$. Here, the first term is $\frac{5}{6}$ and the common ratio is $\frac{5}{6}$. Since $|\frac{5}{6}| < 1$, the sum converges. $\sum_{n=1}^{\infty}\left(\frac{5}{6}\right)^n = \frac{\frac{5}{6}}{1 - \frac{5}{6}} = \frac{\frac{5}{6}}{\frac{1}{6}} = 5$.

Finally, calculate $\ell$: $\ell = \frac{1}{2} \times 5 = \frac{5}{2}$.