97. How many values of n are there such that n! ends in 1998 zeros? (1) Three values of n (2) No values of n (3) Just one (4) Four values of n

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Show that there is no value of n for which ( $2^{n} \times 5^{n}$ ) ends in 5.

For what value of n, 5ⁿ ends with a zero

Find the value of $^{n} C_{0}$ 4 + $4^{2}$ × $\frac{^{n} C_{1}}{2}$ + ......... $\frac{4^{n + 1}}{n + 1}$ × $^{n} C_{n}$

n = 2

5 n - 2

5 n^{2} + 5 n - 2

lim n \to \infty \frac{4 \cdot 5^{n + 1} - 2 \cdot 3^{n}}{6 \cdot 3^{n + 1} + 5^{n}}

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