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Chapter 7: Problem 567

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

Short Answer

Expert verified
The short answer is: (b) \((n+1) 2^n\)

Step by step solution

01

Simplify the Absolute Value and Binomial Coefficients

Since binomial coefficients are always non-negative, we can remove the absolute value signs. Additionally, we know that the binomial coefficient can be represented as: $$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$ Using this representation, let's rewrite the binomial coefficients. Thus, the given expression becomes: $$\frac{n!}{0!(n-0)!}+3\frac{n!}{1!(n-1)!}+5\frac{n!}{2!(n-2)!}+\ldots+(2 n+1)\frac{n!}{n!(n-n)!}$$
02

Simplify the Factorials

Nowimplify the factorials in the expression: $$n!+(3n)+5\frac{n!}{2!}+\ldots+(2 n+1)$$
03

Represent the Expression as a Summation

In order to find a pattern that matches one of the given options, let's rewrite the expression as a summation: $$\sum_{k=0}^{n} (2k+1)\frac{n!}{k!(n-k)!}$$
04

Substitute Binomial Theorem

Recall the binomial theorem: \((a+b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k\). In our case, we want to find an expression that matches the given options. To do this, let's choose \(a = b = 1\): $$(1+1)^n = \sum_{k=0}^n \binom{n}{k} 1^{n-k} 1^{k} = \sum_{k=0}^n \binom{n}{k}$$
05

Find the Derivative and Substitute

Now, let's take the derivative of the above expression with respect to n: $$n(2)^{n-1} = \sum_{k=0}^{n} k\binom{n}{k}$$ Now, notice that our original summation expression has the factors of \(2k+1\) instead of just \(k\). Let's multiply the sum by 2 and add the sum of the odd factors in the binomial coefficients: $$n(2)^{n-1} * 2 + \sum_{k=0}^{n} (2k+1)\binom{n}{k} - 2k\binom{n}{k} = (n+1)2^n$$ After comparing this expression with the given options, we can see that the correct answer is: (b) \((n+1) 2^n\)

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Most popular questions from this chapter

If the ratio of co-efficient of three consecutive terms in expansion of \((1+x)^{n}\) is \(1: 7: 42\) then \(n=\) (a) 35 (b) 45 (c) 55 (d) 65

Constant term in expansion of \(\left[\left\\{(\mathrm{x}+1) /\left\\{(\mathrm{x})^{(2 / 3)}-(\mathrm{x})^{(1 / 3)}+1\right\\}\right\\}-\left\\{(\mathrm{x}-1) /\left\\{\left(\mathrm{x}-(\mathrm{x})^{(1 / 2)}\right\\}\right\\}\right]^{10}\right.\) is \(\ldots\) (a) 190 (b) 200 (c) 210 (d) 220

\(\left[(\sqrt{3}+1)^{6}\right]=\ldots \ldots \ldots ;\) where [] is integer part function. (a) 415 (b) 416 (c) 417 (a) 418

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) \({ }^{50} \mathrm{C}_{44} 3^{6}(2 / 5)^{44}\)

The coefficient of \((1 / x)\) in expansion of \((1+x)^{n}[1+(1 / x)]^{n}\) is (a) \({ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{n}}\) (b) \({ }^{2 \mathrm{n}} \mathrm{C}_{(\mathrm{n}-1)}\) (c) \((1 / 2)\) (d) \({ }^{2 \mathrm{n}} \mathrm{C}_{0}\)
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