What is \(\displaystyle\sum_{r=0}^n\) 2 r C(n, r) equal to ?
Evaluating a Binomial Summation Using the Binomial Theorem
The question asks us to find the value of the summation \(\displaystyle\sum_{r=0}^n\) 2\(^r\) C(n, r), where C(n, r) represents the binomial coefficient \(\binom{n}{r}\).
The given summation can be written as:
\(\displaystyle\sum_{r=0}^n \binom{n}{r} 2^r\)
This summation looks very similar to the expansion formula provided by the Binomial Theorem.
Understanding the Binomial Theorem
The Binomial Theorem states that for any non-negative integer \(n\), the expansion of \((x+y)^n\) is given by:
\((x+y)^n = \displaystyle\sum_{r=0}^n \binom{n}{r} x^{n-r} y^r\)
This formula expands \((x+y)^n\) into a sum of terms, where each term involves a binomial coefficient \(\binom{n}{r}\) and powers of \(x\) and \(y\).
- \(\binom{n}{r}\) is the binomial coefficient, calculated as \(\frac{n!}{r!(n-r)!}\).
- \(r\) is the index of summation, ranging from 0 to \(n\).
- \(x^{n-r}\) is the power of the first term \(x\).
- \(y^r\) is the power of the second term \(y\).
Applying the Binomial Theorem to the Summation
Let's compare the given summation with the Binomial Theorem formula:
- Given Summation: \(\displaystyle\sum_{r=0}^n \binom{n}{r} 2^r\)
- Binomial Theorem Formula: \(\displaystyle\sum_{r=0}^n \binom{n}{r} x^{n-r} y^r\)
We need to find values for \(x\) and \(y\) such that \(\binom{n}{r} x^{n-r} y^r\) matches \(\binom{n}{r} 2^r\) for each term in the sum.
Looking at the term \(\binom{n}{r} 2^r\), we can see that it has the binomial coefficient \(\binom{n}{r}\) and a term \(2^r\). In the Binomial Theorem formula, the term is \(\binom{n}{r} x^{n-r} y^r\).
Comparing the powers of \(y\) and the constant term in the given sum, we can match \(y^r\) with \(2^r\). This suggests \(y = 2\).
Now let's look at the \(x^{n-r}\) part. The given sum \(\binom{n}{r} 2^r\) doesn't explicitly show a term with the power \(n-r\). However, we know that \(1^{n-r} = 1\) for any value of \(n-r\).
So, we can rewrite the given term as \(\binom{n}{r} 1^{n-r} 2^r\).
Now comparing \(\binom{n}{r} 1^{n-r} 2^r\) with \(\binom{n}{r} x^{n-r} y^r\), we can identify:
- \(x^{n-r}\) corresponds to \(1^{n-r}\), so \(x = 1\).
- \(y^r\) corresponds to \(2^r\), so \(y = 2\).
Calculating the Value of the Summation
According to the Binomial Theorem, the sum \(\displaystyle\sum_{r=0}^n \binom{n}{r} x^{n-r} y^r\) is equal to \((x+y)^n\).
Substituting the values \(x = 1\) and \(y = 2\) that we found:
\(\displaystyle\sum_{r=0}^n \binom{n}{r} 1^{n-r} 2^r = (1+2)^n\)
Simplifying the expression on the right side:
\((1+2)^n = 3^n\)
Therefore, the value of the summation \(\displaystyle\sum_{r=0}^n \binom{n}{r} 2^r\) is \(3^n\).
Reviewing the Options
Let's look at the given options:
- Option 1: 2\(^n\)
- Option 2: 3\(^n\)
- Option 3: 2\(^{2n}\)
- Option 4: 3\(^{2n}\)
Our calculated value is \(3^n\), which matches Option 2.
Revision Table: Key Concepts
| Concept | Description | Formula/Example |
|---|---|---|
| Binomial Coefficient | The number of ways to choose \(r\) elements from a set of \(n\) elements, without regard to the order of the elements. Denoted as C(n, r) or \(\binom{n}{r}\). | \(\binom{n}{r} = \frac{n!}{r!(n-r)!}\) |
| Binomial Theorem | A formula that gives the expansion of \((x+y)^n\) for any non-negative integer \(n\). | \((x+y)^n = \displaystyle\sum_{r=0}^n \binom{n}{r} x^{n-r} y^r\) |
| Summation Notation | A concise way to represent the sum of a sequence of terms. | \(\displaystyle\sum_{r=0}^n a_r = a_0 + a_1 + \dots + a_n\) |
Additional Information on Binomial Expansions
The Binomial Theorem is a fundamental result in algebra and combinatorics. It has many applications in various fields, including probability, statistics, and calculus.
Some other important binomial expansions and related identities include:
- The sum of binomial coefficients: \(\displaystyle\sum_{r=0}^n \binom{n}{r} = \binom{n}{0} + \binom{n}{1} + \dots + \binom{n}{n} = (1+1)^n = 2^n\). (This corresponds to setting \(x=1\) and \(y=1\) in the Binomial Theorem).
- Alternating sum of binomial coefficients: \(\displaystyle\sum_{r=0}^n (-1)^r \binom{n}{r} = \binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \dots + (-1)^n \binom{n}{n} = (1-1)^n = 0^n\). (This corresponds to setting \(x=1\) and \(y=-1\). \(0^n\) is 0 for \(n \gt 0\), and 1 for \(n=0\)).
- More generally, for any constant \(c\), \(\displaystyle\sum_{r=0}^n \binom{n}{r} c^r = \displaystyle\sum_{r=0}^n \binom{n}{r} 1^{n-r} c^r = (1+c)^n\). Our problem is a specific case of this identity where \(c=2\).
Recognizing the form of a summation and relating it to the Binomial Theorem is a common technique for evaluating such sums. The key is to identify the \(x\) and \(y\) terms in the general expansion formula that match the terms in the specific summation.
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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