The value Ltn→∞ ∑r=1n ∑t=0r−1 17n⋅ncr⋅rct.5t is equal to

G.E =Ltn→∞17n∑r=1n nCr∑t=0r−1 rCt5t=Ltn→∞17n∑r=1n nCr(1+5)r−5r=Ltn→∞17n∑r=1n nCr6r−∑r=1n Cr5r=Ltn→∞17n7n−6n=1

The value Ltn→∞ ∑r=1n ∑t=0r−1 17n⋅ncr⋅rct.5t is equal to

G.E =Ltn→∞17n∑r=1n nCr∑t=0r−1 rCt5t=Ltn→∞17n∑r=1n nCr(1+5)r−5r=Ltn→∞17n∑r=1n nCr6r−∑r=1n Cr5r=Ltn→∞17n7n−6n=1

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The value $\underset{n \to \infty}{Lt} \sum_{r = 1}^{n} (\sum_{t = 0}^{r - 1} \frac{1}{7^{n}} \cdot n_{c_{r}} \cdot r_{c_{t}} . 5^{t})$ is equal to

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Detailed Solution

$\begin{array}{l} G.E = \underset{n \to \infty}{Lt} \frac{1}{7^{n}} (\sum_{r = 1}^{n}^{n} C_{r} \sum_{t = 0}^{r - 1}^{r} C_{t} 5^{t}) \\ = \underset{n \to \infty}{Lt} \frac{1}{7^{n}} (\sum_{r = 1}^{n}^{n} C_{r} ((1 + 5)^{r} - 5^{r})) \\ = \underset{n \to \infty}{Lt} \frac{1}{7^{n}} (\sum_{r = 1}^{n}^{n} C_{r} 6^{r} - \sum_{r = 1}^{n} C_{r} 5^{r}) \\ = \underset{n \to \infty}{Lt} \frac{1}{7^{n}} (7^{n} - 6^{n}) = 1 \end{array}$