Question

The first three terms in the expansion of a binomial are 1, 10, 40. find it.

Answer 1

Step-by-step explanation:

Expanding the terms of the binomial (a+b)^n, we get:

(a+b)^n = C(n,0)*a^n*b^0 + C(n,1)*a^(n-1)*b^1 + C(n,2)*a^(n-2)*b^2+...

=1*a^n*1 + n*a^(n-1)*b^1 + [n*(n-1)/2]*a^(n-1)*b^2 + ...

= 1 + 10 + 40 + ...

Equating corresponding terms,

1 = 1*a^n*1.........................(1)

10 = n*a^(n-1)*b^1...............(2)

40 = [n*(n-1)/2]*a^(n-1)*b^2..(3)

From (1),

a^n = 1.

This is true:

for a=1 for any n,

for a=-1 for even n,

for n=0. But this could not be a binomial with at least 3 terms.

Assume a=-1.........................(4)

Substituting for a in (2),

10 = n*(-1)^(n-1)*b^1

But this is only true for even n, so n-1 is odd, so

(-1)^(n-1) = -1,

so 10 = n*(-1)*b^1..............(5)

= -n*b

→b = -10 / n.......................(6)

Substituting for a and b in (3),

40 = [n*(n-1)/2]*(-1)^(n-1)*(-10/n)^2

= [n*(n-1)/2]*(-1)*(100/n^2)

= (n-1)*(-50/n)

→-40/50 = (n-1)/n = -4/5

→-40/50 = 5(n-1) = -4n

This has no solution for integer n, so the binomial theorem cannot be applied.

Assume a=1.........................(7)

Substituting for a in (2),

10 = n*1^(n-1)*b^1 = n*1*b = n*b

→b = 10 / n.........................(8)

Substituting for a and b in (3),

40 = [n*(n-1)/2]*1^(n-1)*(10/n)^2

= [n*(n-1)/2]*1*(100/n^2)

= (n-1)*(50/n)

→40/50 = (n-1)/n = 4/5

→5(n-1) = 4n.........................(6)

Solve for n, and you have all required information to complete the expansion of (a+b)^n.