Question

If three successive coefficient in the expansion of 1 x n be 28 56 70 find n

Answer 1

Answer:

It is given that , the  three successive coefficient of $(1\pm x)^n$ is 28, 56 and 70.

Let the successive three terms be, $_{r-1}^{n}\textrm{C},_{r}^{n}\textrm{C},_{r+1}^{n}\textrm{C}$

$_{r-1}^{n}\textrm{C}=28,_{r}^{n}\textrm{C}=56, _{r+1}^{n}\textrm{C}=70 ,\frac{_{r-1}^{n}\textrm{C}}{_{r}^{n}\textrm{C}}=\frac{28}{56}\\\\ \frac{r}{n-r+1}=\frac{1}{2}\\\\ \frac{_{r}^{n}\textrm{C}}{_{r+1}^{n}\textrm{C}}=\frac{56}{70}\\\\ \frac{r+1}{n-r}=\frac{4}{5}$

1.

2 r=n-r +1

3 r -n=1

2.

5 r +5=4 n - 4 r

9 r - 4 n=-5

Equation 2 - 3 equation 1

9 r - 4 n-9 r + 3 n=-5 - 3

-n =- 8

n=8

So, n=8

Answer 2

Step-by-step explanation:

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