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Determine the formula for the coefficient of a term in a binomial expansion.

Mathematics

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Step 1: Determine the formula for the coefficient of a term in a binomial expansion.
The (r+1)(r+1)-th term in the expansion of (a+b)n(a+b)^n is given by Tr+1=(nr)an−rbrT_{r+1} = \binom{n}{r} a^{n-r} b^r.
The coefficient of the (r+1)(r+1)-th term is (nr)\binom{n}{r}.

Step 2: Find the coefficients of the sixth and sixteenth terms.
For the sixth term, r+1=6r+1=6, so r=5r=5. The coefficient is (n5)\binom{n}{5}.
For the sixteenth term, r+1=16r+1=16, so r=15r=15. The coefficient is (n15)\binom{n}{15}.

Step 3: Set the coefficients equal and solve for nn.
Given that the coefficients are equal:

(n5)=(n15)\binom{n}{5} = \binom{n}{15}

According to the property of binomial coefficients, if (nr)=(nk)\binom{n}{r} = \binom{n}{k}, then either r=kr=k or r+k=nr+k=n.
Since 5≠155 \neq 15, we must have 5+15=n5+15=n.

n=5+15n = 5+15 n=20n = 20

The value of nn is:

20\boxed{20}

Step 1: Group the terms to apply the binomial theorem.
Let X=(a+b)X = (a+b). The expression becomes (X−c)3(X-c)^3.

Step 2: Expand (X−c)3(X-c)^3 using the binomial theorem.
The binomial theorem states (A+B)3=A3+3A2B+3AB2+B3(A+B)^3 = A^3 + 3A^2B + 3AB^2 + B^3.
Here, A=XA=X and B=−cB=-c.

(X−c)3=X3+3X2(−c)+3X(−c)2+(−c)3(X-c)^3 = X^3 + 3X^2(-c) + 3X(-c)^2 + (-c)^3 =X3−3cX2+3c2X−c3= X^3 - 3cX^2 + 3c^2X - c^3

Step 3: Substitute X=(a+b)X=(a+b) back into the expanded expression.

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