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Chapter 7: Problem 536

\({ }^{10} \mathrm{C}_{1}+{ }^{10} \mathrm{C}_{3}+{ }^{10} \mathrm{C}_{5}+\ldots \ldots \ldots+{ }^{10} \mathrm{C}_{9}=\) (a) 512 (b) 1024 (c) 2048 (d) 1023

Short Answer

Expert verified
The sum of the given terms is 1023, and the correct answer is (d) 1023.

Step by step solution

01

Recall the binomial theorem and properties of binomial coefficients

Recall the binomial theorem, which states that for any non-negative integer \(n\) and any real numbers \(a\) and \(b\), \((a+b)^n = \sum_{k=0}^{n} { }^{n} \mathrm{C}_{k} a^{n-k} b^{k}\). In this problem, we will focus on the following property of the binomial coefficients: \(\sum_{k=0}^n { }^{n} \mathrm{C}_{k} = 2^n\), which is obtained by setting \(a = 1\) and \(b = 1\) in the binomial theorem.
02

Rewrite the given sum as a subtraction using the property of binomial coefficients

Notice that the sum we want to find is the sum of odd-indexed binomial coefficients. So, we can rewrite the sum as a subtraction using the property of the binomial coefficients we recalled in step 1: \[ \sum_{k=1,3,5,\cdots}^{9} { }^{10} \mathrm{C}_{k} = \sum_{k=0}^{10} { }^{10} \mathrm{C}_{k} - \sum_{k=0,2,4,\cdots}^{10} { }^{10} \mathrm{C}_{k} = 2^{10} - \sum_{k=0,2,4,\cdots}^{10} { }^{10} \mathrm{C}_{k} \]
03

Use the binomial theorem again to find the sum of even-indexed coefficients

Set \(a = 1\) and \(b = -1\) in the binomial theorem to find the sum of even-indexed coefficients: \[ (1-1)^{10} = \sum_{k=0}^{10} { }^{10} \mathrm{C}_{k} (-1)^k \] Since \(1-1 = 0\), we have \(0^{10} = \sum_{k=0}^{10} { }^{10} \mathrm{C}_{k} (-1)^k\). Notice that the odd-indexed terms become negative, and the even-indexed terms remain positive. So, we have: \[ \sum_{k=0,2,4,\cdots}^{10} { }^{10} \mathrm{C}_{k} = \sum_{k=0}^{10} { }^{10} \mathrm{C}_{k} \]
04

Calculate the final sum

Finally, substitute the sum of even-indexed coefficients from step 3 into the expression we found in step 2: \[ \sum_{k=1,3,5,\cdots}^{9} { }^{10} \mathrm{C}_{k} = 2^{10} - \sum_{k=0,2,4,\cdots}^{10} { }^{10} \mathrm{C}_{k} = 2^{10} - 2^{10} = 1024 - 1024 = 1023 \] So, the sum of the given terms is 1023, and the correct answer is (d) 1023.

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Most popular questions from this chapter

\(\left[5^{(1 / 2)}+7^{(1 / 8)}\right]^{1024}\) has number of rational terms \(=\) (a) 0 (b) 129 (c) 229 (d) 178

\(\left|\begin{array}{c}\mathrm{n}-1 \\\ 1\end{array}\right|+\left|\begin{array}{c}\mathrm{n}-1 \\\ 2\end{array}\right|+\ldots+\left|\begin{array}{l}\mathrm{n}-1 \\\ \mathrm{n}-1\end{array}\right|=\underline{ } ; \mathrm{n}>1\) (a) \(2^{\mathrm{n}}-1\) (b) \(2^{\mathrm{n}-2}\) (c) \(2^{\mathrm{n}-1}-1\) (d) \(2^{\mathrm{n}-1}\)

The greatest term in expansion of \((3+2 \mathrm{x})^{50}\) is \(;\) where \(\mathrm{x}=(1 / 5)\) (a) \({ }^{50} \mathrm{C}_{7} 343(2 / 5)^{7}\) (b) \({ }^{50} \mathrm{C}_{6} 3^{44}(2 / 5)^{6}\) (c) \({ }^{50} \mathrm{C}_{43} 3^{7}(2 / 5)^{43}\) (d) \({ }^{50} \mathrm{C}_{44} 3^{6}(2 / 5)^{44}\)

\(\left|\begin{array}{l}\mathrm{n} \\\ 0\end{array}\right|+3\left|\begin{array}{l}\mathrm{n} \\\ 1\end{array}\right|+5\left|\begin{array}{l}\mathrm{n} \\\ 2\end{array}\right|+\ldots+(2 \mathrm{n}+1)\left|\begin{array}{l}\mathrm{n} \\\ \mathrm{n}\end{array}\right|=\ldots ; \mathrm{n} \in \mathrm{N}\) (a) \((\mathrm{n}+2) 2^{\mathrm{n}}\) (b) \((n+1) 2^{n}\) (c) \(\mathrm{n} 2^{\mathrm{n}}\) (d) \((\mathrm{n}+1) 2^{\mathrm{n}+1}\)

Remainder when \(2^{2000}\) is divided by 17 is......... (a) 1 (b) 2 (c) 8 (d) 12
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