Binomial theorem for positive integral Index

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

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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, $\frac{b}{b + c} = \frac{(r + 1) +}{n + 1}$ 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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Binomial theorem for positive integral Index

Remember concepts with our Masterclasses.

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

Suppose a=nCr,b=nCr+1,C=nCr+2 and d=nCr+3 now, aa+b= nCr nCr+nCr+1= nCr n+1Cr+1=r+1n+1Similarly, bb+c=(r+1)+n+1 and cc+d=(r+2)+1n+1 Thus, aa+b+cc+d−2bb+c=0

Ready to Test Your Skills?

free study material

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