Question 36

The value of
$$\sum_{n = 0}^\infty \frac{n_{C_{0}} + n_{C_{1}} + ..... +n_{C_{n}}}{n_{P_{n}}}$$
is

Solution

Numerator is equal to $$2^n$$

Hence $$\sum_{n = 0}^\infty \frac{n_{C_{0}} + n_{C_{1}} + ..... +n_{C_{n}}}{n_{P_{n}}}$$ = $$\sum_{n=0}^{\infty}\frac{2^n}{n!}$$

We know $$e^x=\frac{x^n}{n!}$$

Hence e^2

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