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Chapter 7: Problem 510

Number of terms in expansion of \((\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}})^{10}+(\sqrt{\mathrm{x}}-\sqrt{\mathrm{y}})^{10}\) is \(\ldots \ldots\) (a) 5 (b) 6 (c) 7 (d) 8

Short Answer

Expert verified
The number of terms in the expansion of \((\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}})^{10}+(\sqrt{\mathrm{x}}-\sqrt{\mathrm{y}})^{10}\) is \(6\).

Step by step solution

01

Understand the Binomial Theorem

The binomial theorem is used to find the expansion of a binomial expression raised to some power. It states that for any positive integer n, and any real numbers a and b: \((a+b)^n = \sum_{k=0}^{n} {n \choose k} a^{n-k} b^k\) Where \({n \choose k} = \frac{n!}{k!(n-k)!}\) denotes the binomial coefficient.
02

Expanding \((\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}})^{10}\) using Binomial Theorem

Using the binomial theorem, we can expand the first term as follows: \((\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}})^{10} = \sum_{k=0}^{10} {10 \choose k} (\sqrt{\mathrm{x}})^{10-k} (\sqrt{\mathrm{y}})^k\)
03

Expanding \((\sqrt{\mathrm{x}}-\sqrt{\mathrm{y}})^{10}\) using Binomial Theorem

Similarly, we can expand the second term as follows: \((\sqrt{\mathrm{x}}-\sqrt{\mathrm{y}})^{10} = \sum_{k=0}^{10} {10 \choose k} (\sqrt{\mathrm{x}})^{10-k} (-\sqrt{\mathrm{y}})^k\)
04

Combine the expansions of Step 2 and Step 3

Now, we need to combine the expansions from Step 2 and Step 3: \((\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}})^{10}+(\sqrt{\mathrm{x}}-\sqrt{\mathrm{y}})^{10} = \sum_{k=0}^{10} {10 \choose k} (\sqrt{\mathrm{x}})^{10-k} (\sqrt{\mathrm{y}})^k + \sum_{k=0}^{10} {10 \choose k} (\sqrt{\mathrm{x}})^{10-k} (-\sqrt{\mathrm{y}})^k\) Observe that when k is odd, the terms in the second sum will become negative and cancel out the corresponding terms in the first sum. When k is even, the terms in the second sum will become positive and double the corresponding terms in the first sum. Finally, the combined expression will contain the terms with even k values only. Let's count these terms.
05

Counting the terms with even k values

The possible even k values are 0, 2, 4, 6, 8, and 10. Therefore, there are 6 terms in the expansion of \((\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}})^{10}+(\sqrt{\mathrm{x}}-\sqrt{\mathrm{y}})^{10}\) with even k values. So, the number of terms in the expansion is 6 which corresponds to option (b).

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Most popular questions from this chapter

If \(\mathrm{x}\) is so small that terms with \(\mathrm{x}^{3}\) and higher powers of \(\mathrm{x}\) may be neglected then \(\left[\left\\{(1+\mathrm{x})^{(3 / 2)}-\\{1+(\mathrm{x} / 2)\\}^{3}\right\\} /\left\\{(1-\mathrm{x})^{(1 / 2)}\right\\}\right]\) may be approximated as \(\ldots \ldots \ldots\) (a) \([(-3) / 8] \mathrm{x}^{2}\) (b) \((1 / 2) \mathrm{x}-(3 / 8) \mathrm{x}^{2}\) (c) \(1-(3 / 8) x^{2}\) (d) \(3 x+(3 / 8) x^{2}\)

\(\left[(\sqrt{2}+1)^{8}\right]=\ldots \ldots \ldots ;\) where [] is integer part function. (a) 1151 (b) 1152 (c) 1153 (a) 1154

Number of irrational terms in expansion of \(\left[4^{(1 / 5)}+7^{(1 / 10)}\right]^{45}=\) (a) 40 (b) 5 (c) 41 (d) 8

Sum of Co-efficient of last 15 terms in expansion of \((1+\mathrm{x})^{29}\) is (a) \(2^{15}\) (b) \(2^{30}\) (c) \(2^{29}\) (d) \(2^{28}\)

It \(\mathrm{A}\) and \(\mathrm{B}\) are coefficients of \(\mathrm{x}^{\mathrm{r}}\) and \(\mathrm{x}^{\mathrm{n}-\mathrm{r}}\) respectively in expansion of \((1+x)^{n}\) then \(=\) (a) \(\mathrm{A}+\mathrm{B}=\mathrm{n}\) (b) \(\mathrm{A}=\mathrm{B}\) (c) \(\mathrm{A}+\mathrm{B}=2^{\mathrm{n}}\) (d) \(A-B=2^{n}\)
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