The value of E=(1+17)1+1721+173â¦1+1719(1+19)1+1921+193â¦1+1917 is

The value of $E = \frac{(1 + 17) (1 + \frac{17}{2}) (1 + \frac{17}{3}) â¦ (1 + \frac{17}{19})}{(1 + 19) (1 + \frac{19}{2}) (1 + \frac{19}{3}) â¦ (1 + \frac{19}{17})}$ is

A
1
B
$^{36} C_{17}$
C
$\frac{2}{19}$
D
$^{36} C_{18}$

Solution:

We have

$(1 + k) (1 + \frac{k}{2}) (1 + \frac{k}{3}) â¦ (1 + \frac{k}{n})$

$\begin{array}{l} = \frac{(1 + k) (2 + k) (3 + k) â¦ (n + k)}{(2) (3) â¦ (n)} = \frac{(n + k)!}{k! n!} \\ =^{n + k} C_{k} \end{array}$

Thus, both the numerator and the denominator of $E$ equals

$36_{C_{17}} = 36_{C_{19}} â$

$â´ E = 1 .$

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