The binomial coefficients of the expansion (a+b)^m are the consecutive tems of an arithmetical sequence.The term of expansion that is 21 is having b^5. a =square root( 2^lg(10-3^x))...

The binomial coefficients of the expansion (a+b)^m are the consecutive tems of an arithmetical sequence.

The term of expansion that is 21 is having b^5.

a =square root( 2^lg(10-3^x))

b=[2^(x-2)*lg3]^1/5

What is x?

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We'll start using the condition from enunciation that the binomial coefficients of the given expansion are in arithmetical progression.

Since it is not indicated what are the terms, we'll suppose that they are the 2nd, 3rd and the 4th terms.

C(m,2) = [C(m,1) + C(m,3)]/2

We'll re-write the condition above:

m(m-1) = m + m(m-1)(m-2)/6

We'll multiply by 6 all the terms:

6m(m-1) = 6m + m(m-1)(m-2)

6m(m-1) = m[6 + (m-1)(m-2)]

We'll divide by m:

6m - 6 = m^2 - 3m + 8

We'll move all terms to one side and we'll combine them:

m^2 - 9m + 14 = 0

We'll apply the quadratic formula:

m1 = (9+sqrt25)/2

m1 = (9+5)/2

m1 = 7

m2 = (9-5)/2

m2 = 2

But it is impossible for the value of m to be smaller than the value of k, that is 5, from the condition from enunciation, that one term is containing b^5.

Putting m = 7, we'll write the formula of the genertal term of the expansion:

T6 = C(7,5)*a62*b^5

21 = 7(7-1)/2*2^lg(10-3^x)*2^(x-2)*lg3

We'll divide by 21 both sides:

2^lg(10-3^x)*2^(x-2)*lg3 = 1

Since the bases are matching, we'll add the superscripts:

2^[lg(10-3^x)+(x-2)*lg3] = 2^0

[lg(10-3^x)+(x-2)*lg3] = 0

We'll use the power property:

[lg(10-3^x)+lg3^(x-2)] = 0

lg(10-3^x) = -lg3^(x-2)

lg(10-3^x) = lg3^-(x-2)

Since the bases are matching, we'll use one to one property:

10-3^x = 3^-(x-2)

10-3^x = 9/3^x

We'll substitute 3^x by t:

10 - t = 9/t

10t - t^2 - 9 = 0

t^2 - 10t + 9 = 0

t1 = 9 and t2 = 1

3^x = 9

x = 2

t2 = 1

x = 0

The requested values of x are {0 ; 2}.

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