Write down the general term in the binomial expansion of (x+5x)n\left(x + \frac{5}{x}\right)^n. Given that there is a constant term in the expansion of (x+5x)n\left(x + \frac{5}{x}\right)^n, show that nn must be an even number. Find the term in x6x^6 in the expansion of (x+5x)10\left(x + \frac{5}{x}\right)^{10}. Hence or otherwise, find the value of aa, given that the coefficient of x6x^6 is 43875 in the expansion of (ax2−1)(x+5x)10(ax^2 - 1)\left(x + \frac{5}{x}\right)^{10}.

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Worked solution

7(i)

1. Write down the general term formula for the given binomial

The general term Tr+1T_{r+1} in the expansion of (a+b)n(a+b)^n is (nr)an−rbr\binom{n}{r} a^{n-r} b^r.
For (x+5x)n\left(x + \frac{5}{x}\right)^n, we have:
Tr+1=(nr)xn−r(5x)rT_{r+1} = \binom{n}{r} x^{n-r} \left(\frac{5}{x}\right)^r

2. Simplify the general term

Tr+1=(nr)xn−r5rxrT_{r+1} = \binom{n}{r} x^{n-r} \frac{5^r}{x^r}
Tr+1=(nr)5rxn−2rT_{r+1} = \binom{n}{r} 5^r x^{n-2r}

Answer

Tr+1=(nr)5rxn−2rT_{r+1} = \binom{n}{r} 5^r x^{n-2r}

7(ii)

1. Equate the power of $x$ in the general term to 0 for a constant term

From part (i), the general term is Tr+1=(nr)xn−r(5x)r=(nr)5rxn−2rT_{r+1} = \binom{n}{r} x^{n-r} \left(\frac{5}{x}\right)^r = \binom{n}{r} 5^r x^{n-2r}.
For a constant term, the power of xx must be zero:
n−2r=0n - 2r = 0

2. Deduce that $n$ must be even

n=2rn = 2r
Since rr is an integer (0⩽r⩽n0 \leqslant r \leqslant n), 2r2r is an even number. Therefore, nn must be an even number.

Answer

Shown.

7(iv)

1. Identify the terms from the binomial expansion needed to form $x^6$ in the product

The term in x6x^6 in the expansion of (ax2−1)(x+5x)10(ax^2 - 1)\left(x + \frac{5}{x}\right)^{10} is obtained by multiplying ax2ax^2 by the term in x4x^4 from (x+5x)10\left(x + \frac{5}{x}\right)^{10}, and −1-1 by the term in x6x^6 from (x+5x)10\left(x + \frac{5}{x}\right)^{10}.

2. Find the term in $x^4$ and the term in $x^6$ from the binomial expansion

The general term is Tr+1=(10r)x10−r(5x)r=(10r)5rx10−2rT_{r+1} = \binom{10}{r} x^{10-r} \left(\frac{5}{x}\right)^r = \binom{10}{r} 5^r x^{10-2r}.

For the term in x6x^6 (from part iii):
10−2r=6  ⟹  r=210 - 2r = 6 \implies r = 2
Term in x6=(102)52x6=45(25)x6=1125x6x^6 = \binom{10}{2} 5^2 x^6 = 45(25)x^6 = 1125x^6

For the term in x4x^4:
10−2r=4  ⟹  r=310 - 2r = 4 \implies r = 3
Term in x4=(103)53x4=120(125)x4=15000x4x^4 = \binom{10}{3} 5^3 x^4 = 120(125)x^4 = 15000x^4

3. Form an equation using the given coefficient and solve for $a$

Coefficient of x6x^6 in the product:
a(15000)−1(1125)=43875a(15000) - 1(1125) = 43875
15000a−1125=4387515000a - 1125 = 43875
15000a=4500015000a = 45000
a=3a = 3

Answer

a=3a = 3

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