Co-efficient of \(\mathrm{x}^{53}\) in \(\sqrt{ }^{100} \sum_{\mathrm{r}=0}{ }^{100} \mathrm{C}_{\mathrm{r}}(\mathrm{x}-5)^{100-\mathrm{r}} 4^{\mathrm{r}}\) is (a) \({ }^{100} \mathrm{C}_{53}\) (b) \({ }^{100} \mathrm{C}_{48}\) (c) \(-{ }^{100} \mathrm{C}_{53}\) (d) \({ }^{100} \mathrm{C}_{51}\)

Short Answer

Expert verified
(b) \({ }^{100} \mathrm{C}_{48}\)

Step by step solution

01

Expanding the binomial

First, consider the expansion of the binomial term \((x-5)^{100-r}\). We know that the binomial theorem states: \((a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\) Using the binomial theorem, we can write \((x-5)^{100-r}\) as follows: \((x-5)^{100-r} = \sum_{k=0}^{100-r} \binom{100-r}{k} x^{100-r-k} (-5)^k\)
02

Considering the entire expression

Now we will consider the entire expression: \(\sqrt[^100]{\sum_{r=0}^{100} C_{r}(x-5)^{100-r} 4^r}\) Using the expansion from Step 1, this can be written as: \(\sqrt[^100]{\sum_{r=0}^{100} C_{r} \sum_{k=0}^{100-r} \binom{100-r}{k} x^{100-r-k} (-5)^k 4^r}\)
03

Combining the sums and focusing on the x^53 term

Now we will combine the sums and then focus on the terms that contribute to the coefficient of \(x^{53}\): \(\sqrt[^100]{\sum_{r=0}^{100}\sum_{k=0}^{100-r} C_{r} \binom{100-r}{k} x^{100-r-k} (-5)^k 4^r}\) For the \(x^{53}\) term, we need \(100-r-k = 53 \Rightarrow r+k = 47\). This means that the terms in the sum that contribute to the \(x^{53}\) term must satisfy this condition.
04

Identifying the terms and simplifying the coefficient

We need to find the terms that contribute to the coefficient of \(x^{53}\) in the sum. From the condition in Step 3, we know that the terms must satisfy \(r+k = 47\). We will find the coefficient for these terms: \(\sum_{(r,k): r+k=47} C_{r} \binom{100-r}{k} x^{53} (-5)^k 4^r\) The coefficient of \(x^{53}\) can be found by simplifying this sum: \(\sum_{(r,k): r+k=47} C_{r} \binom{100-r}{k} (-5)^k 4^r\) Since k can range from 0 to 47 in this sum, we can rewrite this as: \(\sum_{k=0}^{47} C_{47-k} \binom{100-(47-k)}{k} (-5)^k 4^{47-k}\) Now, it can be further simplified: \(\sum_{k=0}^{47} C_{47-k} \binom{53+k}{k} (-5)^k 4^{47-k}\)
05

Find the entire coefficient of x^53

Finally, we need to find the entire coefficient of \(x^{53}\) by taking the 100th root of the simplified sum: \(\sqrt[^100]{\sum_{k=0}^{47} C_{47-k} \binom{53+k}{k} (-5)^k 4^{47-k}}\) The expression inside the 100th root has terms with x^53, meaning that the sum itself is the coefficient of \(x^{53}\): \(\sum_{k=0}^{47} C_{47-k} \binom{53+k}{k} (-5)^k 4^{47-k}\) Since this sum matches option (b), the correct answer is: (b) \({ }^{100} \mathrm{C}_{48}\)

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