Question

The first three terms in the binomial expansion of ( ) n x y + are 1, 56 and 1372 respectively . Find x y and n

Answer 1

x = 1  , y = 7  , n = 8  if  first three terms in the binomial are 1, 56 and 1372

Step-by-step explanation:

(x + y)ⁿ

= xⁿ  + ⁿC₁xⁿ⁻¹y  + ⁿC₂xⁿ⁻²y² +............................

First Term xⁿ = 1

=> x = 1

ⁿC₁xⁿ⁻¹y  = 56

ⁿC₂xⁿ⁻²y² = 1372

=> ⁿC₂xⁿ⁻²y² / ⁿC₁xⁿ⁻¹y   = 1372 /56

=> (n - 1) y/ 2x   = 49/2

=>  (n - 1) y/x  = 49

using x = 1

=> (n - 1) y  = 49

7 * 7

n = 8  , y = 7

(1  + 7)⁸

1 + 8 * 7  + 28 * 7² +......

x = 1  , y = 7  , n = 8

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Answer 2

x=1,n=8,y=7

Step-by-step explanation:

Binomial expansion

$(x+y)^n=nC_0x^n+nC_1x^{n-1}y+nC_2x^{n-2}y^2+....+nC_ny^n$

According to question

$nC_0x^n=1$

$nC_1x^{n-1}y=56$

$nC_2x^{n-2}y^2=1372$

Combination formula:

$nC_r=\frac{n!}{r!(n-r)!}$

By using combination formula

$x^n=1$

$x^n=1$

We get

x=1

$\frac{n!}{(n-1)!}x^{n-1}y=56$

$\frac{n(n-1)!}{(n-1)!}x^{n-1}y=56$

Substitute the value of x

$ny=56$

$\frac{n!}{2!(n-2)!}x^{n-2}y^2=1372$

$\frac{\frac{n!}{(n-1)!}x^{n-1}y}{\frac{n!}{2!(n-2)!}x^{n-2}y^2}=\frac{56}{1372}$

$\frac{2(n-2)!}{(n-1)(n-2)!}\times \frac{x}{y}=\frac{2}{49}$

Substitute the value of x

$\frac{1}{y(n-1)}=\frac{1}{49}$

$(n-1)y=49$

$ny-y=49$

Substitute the value of ny

$56-y=49$

$y=56-49=7$

y=7

Substitute the value of y

$n(7)=56$

$n=\frac{56}{7}=8$

n=8

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