Problem 52
Question
The sum of the series \(\sum_{r=0}^{10}{\underline{\phantom{xx}}}^{20} C_{r}\) is (A) \(2^{19}-\frac{1}{2} \cdot{ }^{20} C_{10}\) (B) \(2^{19}+\frac{1}{2} \cdot{ }^{20} C_{10}\) (C) \(2^{19}\) (D) \(2^{20}\)
Step-by-Step Solution
Verified Answer
The sum of the series is \(2^{19}\), which is option (C).
1Step 1: Understand the Series
The given series is \( \sum_{r=0}^{10} \binom{20}{r} \). This means we are summing the binomial coefficients \( \binom{20}{0}, \binom{20}{1}, \ldots, \binom{20}{10} \).
2Step 2: Use the Binomial Theorem
According to the binomial theorem, the sum of all binomial coefficients for a given number \( n \) is \( 2^n \). Thus, \( \sum_{r=0}^{20} \binom{20}{r} = 2^{20} \).
3Step 3: Compute the Complement Fraction
Notice that \( \sum_{r=0}^{10} \binom{20}{r} \) represents only half the terms of the full binomial expansion. This is symmetric such that \( \sum_{r=0}^{10} \binom{20}{r} = \sum_{r=11}^{20} \binom{20}{r} \). Thus, \( 2 \sum_{r=0}^{10} \binom{20}{r} = 2^{20} \).
4Step 4: Solve for the Desired Sum
From Step 3, \( \sum_{r=0}^{10} \binom{20}{r} = \frac{2^{20}}{2} = 2^{19} \). It matches option (C).
Key Concepts
Binomial CoefficientsSymmetric ExpansionBinomial Expansion
Binomial Coefficients
Binomial coefficients are essential components in the field of combinatorics and are heavily used in the binomial theorem for expansions. A binomial coefficient is represented as \( \binom{n}{r} \), which is read as "n choose r". It tells us the number of ways to choose \( r \) elements from a set of \( n \) elements. The formula to calculate a binomial coefficient is:\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]Here, \( n! \) denotes the factorial of \( n \), which is the product of all positive integers up to \( n \). Binomial coefficients are symmetrical, meaning \( \binom{n}{r} = \binom{n}{n-r} \), which plays a key role in simplifying calculations within the binomial expansion.
Binomial coefficients appear in the expansion of a binomial expression raised to a power, such as \((a+b)^n \), where they form Pascal's triangle, a triangular array of the coefficients calculated by a recursive method. Understanding binomial coefficients allows one to find specific terms in a polynomial expansion without fully expanding it.
Binomial coefficients appear in the expansion of a binomial expression raised to a power, such as \((a+b)^n \), where they form Pascal's triangle, a triangular array of the coefficients calculated by a recursive method. Understanding binomial coefficients allows one to find specific terms in a polynomial expansion without fully expanding it.
Symmetric Expansion
Symmetric expansion relies on the property of binomial coefficients being symmetrical. This symmetry is generally expressed as \( \binom{n}{r} = \binom{n}{n-r} \). In a full binomial expansion like \((a+b)^n\), each term is mirrored around the midpoint of the expansion.
For instance, in the binomial expansion of \((a + b)^{20} \), the terms from \( \binom{20}{0} \) to \( \binom{20}{10} \) mirror those from \( \binom{20}{20} \) to \( \binom{20}{11} \). This means that both halves of the expansion are identical when summed. If one calculates only half the terms, they can determine the sum of the other half by leveraging this symmetry.
The given exercise uses this property to simplify the calculation of the sum for the first half of the series \( \sum_{r=0}^{10} \binom{20}{r} \), by showing that it is equivalent to the sum from \( r=11 \) to \( r=20 \). This leads to a simplified solution, as calculating half the sum directly gives you half of the entire series.
For instance, in the binomial expansion of \((a + b)^{20} \), the terms from \( \binom{20}{0} \) to \( \binom{20}{10} \) mirror those from \( \binom{20}{20} \) to \( \binom{20}{11} \). This means that both halves of the expansion are identical when summed. If one calculates only half the terms, they can determine the sum of the other half by leveraging this symmetry.
The given exercise uses this property to simplify the calculation of the sum for the first half of the series \( \sum_{r=0}^{10} \binom{20}{r} \), by showing that it is equivalent to the sum from \( r=11 \) to \( r=20 \). This leads to a simplified solution, as calculating half the sum directly gives you half of the entire series.
Binomial Expansion
The binomial expansion refers to the expansion of a power of a binomial expression based on the binomial theorem. The binomial theorem states that:\[ (a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]Each term in this expansion has a co-efficient \( \binom{n}{k} \), a power of \( a \), and a power of \( b \). These terms outline how one can expand a simple binomial into a sum of terms of different powers of \( a \) and \( b \).
In practical scenarios, such as the given exercise, binomial expansion allows the calculation of sums involving binomial coefficients without expanding each term individually. For example, calculating \( \sum_{r=0}^{20} \binom{20}{r} \) involves using the property of the binomial expansion rather than manually working through each coefficient.
Moreover, the expansion is used to determine specific probabilities and outcomes in various probability-related questions, as each binomial expansion expresses the likelihood of different combinations happening when there are two possible outcomes. Comprehending how the binomial expansion works can simplify complex problems in both algebra and probability.
In practical scenarios, such as the given exercise, binomial expansion allows the calculation of sums involving binomial coefficients without expanding each term individually. For example, calculating \( \sum_{r=0}^{20} \binom{20}{r} \) involves using the property of the binomial expansion rather than manually working through each coefficient.
Moreover, the expansion is used to determine specific probabilities and outcomes in various probability-related questions, as each binomial expansion expresses the likelihood of different combinations happening when there are two possible outcomes. Comprehending how the binomial expansion works can simplify complex problems in both algebra and probability.
Other exercises in this chapter
Problem 50
Larger of \(99^{50}+100^{50}\) and \(101^{50}\) is (A) \(101^{50}\) (B) \(99^{50}+100^{50}\) (C) both are equal (D) none of these
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View solution Problem 53
\({ }^{\mathrm{n}+1} \mathrm{C}_{2}+2\left[{ }^{2} \mathrm{C}_{2}+{ }^{3} \mathrm{C}_{2}+{ }^{4} \mathrm{C}_{2}+\ldots+{ }^{\mathrm{n}} C_{2}\right]=\) (A) \(\f
View solution Problem 54
If \(A={ }^{2 n} C_{0} \cdot{ }^{2 n} C_{1}+{ }^{2 n} C_{1}{\underline{\phantom{xx}}}^{2 \mathrm{n}-1} C_{1}+{ }^{2 n} C_{2}{\underline{\phantom{xx}}}^{2 \mathrm{n}-2} C_{1}+\ldots\), then \(A\) is (A) 0 (B
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