What is the sum of the coefficients of first and last terms in the expansion of (1 + x) 2n , where n is a natural number?
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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
| Coefficient Formula | Meaning | Property |
|---|---|---|
| \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\) | Number of ways to choose \(k\) items from a set of \(n\) distinct items. | \(\binom{n}{k} = \binom{n}{n-k}\) |
| \(\binom{n}{0}\) | Number of ways to choose 0 items from \(n\). | \(\binom{n}{0} = 1\) |
| \(\binom{n}{n}\) | Number of ways to choose \(n\) items from \(n\). | \(\binom{n}{n} = 1\) |
| \(\binom{n}{1}\) | Number of ways to choose 1 item from \(n\). | \(\binom{n}{1} = n\) |
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.
Important Questions from Binomial Expansion
If $x = \frac{1}{4}$, then the greatest term in the expansion of $(2 + 3x)^{15}$ will be
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