Understanding the Binomial Expansion

The question asks for the sum of the coefficients of the first and last terms in the expansion of \((1 + x)^{2n}\), where \(n\) is a natural number. To solve this, we need to recall the Binomial Theorem.

Applying the Binomial Theorem

The Binomial Theorem states that for any natural number \(m\), the expansion of \((a+b)^m\) is given by:

\[(a+b)^m = \sum_{k=0}^m \binom{m}{k} a^{m-k} b^k\]

where \(\binom{m}{k}\) are the binomial coefficients, calculated as \(\frac{m!}{k!(m-k)!}\).

In our case, the expression is \((1 + x)^{2n}\). Here, \(a=1\), \(b=x\), and the exponent is \(m=2n\). Substituting these values into the binomial theorem formula, we get:

\[(1 + x)^{2n} = \sum_{k=0}^{2n} \binom{2n}{k} (1)^{2n-k} (x)^k\]

Since \(1\) raised to any power is \(1\), the expansion simplifies to:

\[(1 + x)^{2n} = \sum_{k=0}^{2n} \binom{2n}{k} x^k\]

Let's write out the terms of this expansion:

\[(1 + x)^{2n} = \binom{2n}{0} x^0 + \binom{2n}{1} x^1 + \binom{2n}{2} x^2 + \dots + \binom{2n}{2n-1} x^{2n-1} + \binom{2n}{2n} x^{2n}\]

The general term in the expansion is \(T_{k+1} = \binom{2n}{k} x^k\). The coefficient of this term is \(\binom{2n}{k}\).

Identifying the First and Last Terms

The expansion starts with \(k=0\) and ends with \(k=2n\).

• The first term corresponds to \(k=0\): \(T_1 = \binom{2n}{0} x^0 = \binom{2n}{0} \cdot 1 = \binom{2n}{0}\).
• The last term corresponds to \(k=2n\): \(T_{2n+1} = \binom{2n}{2n} x^{2n}\).

Finding the Coefficients

The coefficient of the first term (\(T_1\)) is \(\binom{2n}{0}\).

Recall that \(\binom{m}{0} = 1\) for any non-negative integer \(m\). Therefore, the coefficient of the first term is \(\binom{2n}{0} = 1\).

The coefficient of the last term (\(T_{2n+1}\)) is \(\binom{2n}{2n}\).

Recall that \(\binom{m}{m} = 1\) for any non-negative integer \(m\). Therefore, the coefficient of the last term is \(\binom{2n}{2n} = 1\).

Calculating the Sum of Coefficients

We are asked for the sum of the coefficients of the first and last terms. Sum = (Coefficient of first term) + (Coefficient of last term)

Sum = \(\binom{2n}{0} + \binom{2n}{2n}\)

Sum = \(1 + 1\)

Sum = \(2\)

Thus, the sum of the coefficients of the first and last terms in the expansion of \((1 + x)^{2n}\) is 2.

Revision Table: Binomial Coefficients

Additional Information on Binomial Expansion

The sum of all coefficients in the expansion of \((a+b)^m\) is found by setting \(a=1\) and \(b=1\). For \((1+x)^{2n}\), the sum of all coefficients is obtained by setting \(x=1\), which gives \((1+1)^{2n} = 2^{2n}\).

The binomial coefficients \(\binom{2n}{k}\) are symmetric, meaning \(\binom{2n}{k} = \binom{2n}{2n-k}\). This property is evident in the coefficients of the first and last terms: \(\binom{2n}{0} = \binom{2n}{2n}\).

The expansion \((1+x)^{2n}\) has \(2n+1\) terms.