Question

if the consecutive coefficient in the expansion of (1+x)^n be 56.70 and 56. find n and the position of the coefficients.​

Answer 1

Given Question :-

If the three consecutive coefficient in the expansion of (1+x)^n be 56, 70 and 56. Find n and the position of the coefficients.

Answer :-

Let three consecutive terms be r, (r + 1) and (r + 2).

We know that

$\sf \: In \: expansion \: of \: {(1 + x)}^{n} , \: the \: coefficient \: of \:$

$\sf \: T_r \: = \:^n C_{r - 1}$

and

Also,

$\rm :\longmapsto\:\dfrac{\:^n C_x}{\:^n C_y} = \dfrac{n - y}{x} \: \: \: if \: x - y = 1$

Now,

$\sf \: Coefficient \: of \: T_r = \:^n C_{r - 1}$

$\sf \: Coefficient \: of \: T_{r + 1} = \:^n C_{r}$

$\sf \: Coefficient \: of \: T_{r + 2} = \:^n C_{r + 1}$

Now,

• According to statement,

$\rm :\longmapsto\:\:^n C_{r - 1} = 56 - - (1)$

$\rm :\longmapsto\:\:^n C_r = 70 - - - (2)$

$\rm :\longmapsto\:\:^n C_{r + 1} = 56 - - - (3)$

Now,

• Divide equation (2) by equation (1), we get

$\rm :\longmapsto\:\dfrac{\:^n C_r}{\:^n C_{r - 1}} = \dfrac{70}{56}$

$\rm :\longmapsto\:\dfrac{n - (r - 1)}{r} = \dfrac{5}{4}$

$\rm :\longmapsto\:\dfrac{n - r + 1}{r} = \dfrac{5}{4}$

$\rm :\longmapsto\:4n - 4r + 4 = 5r$

$\rm :\implies\:9r - 4n = 4 - - - (4)$

Now,

• Divide equation (3) by equation (2), we get

$\rm :\longmapsto\:\dfrac{\:^n C_{r + 1}}{\:^n C_r} = \dfrac{56}{70}$

$\rm :\longmapsto\:\dfrac{n - r}{r + 1} = \dfrac{4}{5}$

$\rm :\longmapsto\:5n - 5r = 4r + 4$

$\bf\implies \:5n - 9r = 4 - - - (5)$

• On Adding equation (4) and equation (5), we get

$\bf\implies \:n \: = \: 8$

• On substituting n = 8, in equation (4), we get

$\rm :\longmapsto\:9r - 4 \times 8 = 4$

$\rm :\longmapsto\:9r - 32 = 4$

$\rm :\longmapsto\:9r = 36$

$\bf\implies \:r = 4$

$\bf\implies \:3 \: consecutive \: terms \: are \: 4, \: 5 \: and \: 6$

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Additional Information :-

$\boxed{ \bf{ \: \:^n C_x = \:^n C_y\rm \implies\:x = y \: or \: n = x + y}}$

$\boxed{ \bf{ \: \:^n C_r + \:^n C_{r - 1} = \:^{n + 1} C_r}}$

$\boxed{ \bf{ \:\dfrac{\:^n C_r}{\:^n C_{r - 1}} = \dfrac{n - r + 1}{r}}}$

$\boxed{ \bf{ \: Coefficient \: of \: {x}^{r} \:in \: expansion \: of \: {(1 + x)}^{n} \: is \: \:^n C_r}}$