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Chapter 7: Problem 520

The \(11^{\text {th }}\) term from last, in expansion of \(\left[2 \mathrm{x}+\left(1 / \mathrm{x}^{2}\right)\right]^{25}\) is (a) \(-{ }^{25} \mathrm{C}_{15}\left(2^{10} / \mathrm{x}^{20}\right)\) (b) \(+{ }^{25} \mathrm{C}_{15}\left(2^{10} / \mathrm{x}^{20}\right]\) (c) \(-{ }^{25} \mathrm{C}_{14}\left(2^{11} / \mathrm{x}^{11}\right)\) (d) \({ }^{25} \mathrm{C}_{14}\left(2^{11} / \mathrm{x}^{11}\right)\)

Short Answer

Expert verified
The short answer is: \({ }^{25}C_{14}\ 2^{11}x^{-11}\)

Step by step solution

01

Using binomial theorem to find general term

We can expand the given expression using the binomial theorem. For any positive integer n, \((a+b)^n\) can be expanded as \[\sum_{k=0}^{n}\ ^nC_k\ a^{(n-k)}\ b^k\] where \(_{n} C_{k}\) represents the binomial coefficients. Now apply this to the given expression \(\left[2x+\left(\frac{1}{x^2}\right)\right]^{25}\): Then, the general term in the expansion will be \[^{25}C_k\ (2x)^{(25-k)}\left(\frac{1}{x^2}\right)^k\]
02

Simplifying the general term

Next, we will rewrite and simplify the general term based on the binomial theorem: \[^{25}C_k\ (2^{25-k}x^{(25-k)})\left(\frac{1}{x^{2k}}\right)\] which simplifies to \[^{25}C_k\ 2^{25-k}x^{25-3k}\]
03

Finding the 11th term from the last

To find the 11th term from the last, we need to find the corresponding term from the beginning. Here, the total number of terms in the expansion is 26 (from \(k=0\) to \(k=25\), inclusive), so the 11th term from the last is equivalent to the 16th term from the beginning (26 - 11 = 15, means k=14). Hence, substitute \(k=14\) in the simplified general term: \[^{25}C_{14}\ 2^{25-14}\ x^{25-3 \times 14} = { }^{25}C_{14}\ 2^{11}x^{-11}\]
04

Comparing and choosing the correct option

We now have the 11th term from the last in the expanded form: \[^{25}C_{14}\ 2^{11}x^{-11}\] Compare this with the given options to identify the correct one: (a) \(-{ }^{25}C_{15}\left(2^{10}\ /x^{20}\right)\): incorrect (b) \(+{ }^{25}C_{15}\left(2^{10}\ /x^{20}\right)\): incorrect (c) \(-{ }^{25}C_{14}\left(2^{11}\ /x^{11}\right)\): incorrect (d) \(+{ }^{25}C_{14}\left(2^{11}\ /x^{11}\right)\): correct The correct answer is (d) \(+{ }^{25}C_{14}\left(2^{11}\ /x^{11}\right)\).

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Most popular questions from this chapter

\(17^{\text {th }}\) term and \(18^{\text {th }}\) term are equal in expansion of \((2+x)^{40}\) then \(\mathrm{x}=\) (a) \((17 / 24)\) (b) \((17 / 12)\) (c) \((34 / 13)\) (d) \((34 / 15)\)

If rth term in expansion of \([x+(1 / 2 x)]^{12}\) is constant then \(r\) (a) 5 (b) 6 (c) 7 (d) 8

If sum of Even terms are denoted by \(\mathrm{E}\) and sum of odd terms are denoted by \(\mathrm{O}\) in expansion of \((\mathrm{x}+\mathrm{a})^{\mathrm{n}}\) then \(\mathrm{O}^{2}-\mathrm{E}^{2}=\) (a) \(x^{2}-a^{2}\) (b) \(\left(\mathrm{x}^{2}-\mathrm{a}^{2}\right)^{\mathrm{n}}\) (c) \(x^{2 n}-a^{2 n}\) (b) None of these

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) \({ }^{100} \mathrm{C}_{51}\)

\(\left[\left(a^{1 / 3} / b^{1 / 6}\right)+\left(b^{1 / 2} / a^{1 / 6}\right)\right]^{21}\) has same power of a and \(b\) for \((\mathrm{r}+1)\) th term then \(\mathrm{r}=\) (a) 8 (b) 9 (c) 10 (d) 11
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