Let a, b, c, d be any four consecutive coefficients in the binomial expansion of (1 + x)n,thenaa+b+cc+d−2bb+c is equal to

Let a, b, c, d be any four consecutive coefficients in the binomial expansion of ${(1 + x)}^{n}$ ,then
$\frac{a}{a + b} + \frac{c}{c + d} - 2 (\frac{b}{b + c})$ is equal to

A
n
B
1
C
0
D
2n

Solution:

Suppose $a =^{n} C_{r}, b =^{n} C_{r + 1}, C =^{n} C_{r + 2}$ and $d =^{n} C_{r + 3}$ Now

$\frac{a}{a + b} = \frac{^{n} C_{r}}{^{n} C_{r} +^{n} C_{r + 1}} = \frac{^{n} C_{r}}{^{n + 1} C_{r + 1}} = \frac{r + 1}{n + 1}$

Similarly, $\frac{b}{b + c} = \frac{(r + 1) + 1}{n + 1}$ and $\frac{c}{c + d} = \frac{(r + 2) + 1}{n + 1}$

Thus, $\frac{a}{a + b} + \frac{c}{c + d} - 2 (\frac{b}{b + c}) = 0$

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