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NDA 2024
Exam
NDA
Year
2024
Subject
Mathematics
Chapter
Binomial Theorem

NDA 2024 · Mathematics · Binomial Theorem

Paragraph: Let (8+3 7 )^ 20 =U+V and (8-3 7 )^ 20 =W , where U is...

Paragraph: Let (8+37)20=U+V(8+3 \sqrt{7})^{20}=\mathrm{U}+\mathrm{V} and (8−37)20=W(8-3 \sqrt{7})^{20}=\mathrm{W}, where U is an integer and 0<V<10 < \mathrm{V} < 1.
Question: What is the value of (U+V)W(\mathrm{U}+\mathrm{V}) \mathrm{W} ?

Options

  1. A12\frac{1}{2}
  2. B11
  3. C32\frac{3}{2}
  4. D22

Correct answer

B. 11

Step-by-step solution

We have, (8+37)20=U+V(8+3 \sqrt{7})^{20}=\mathrm{U}+\mathrm{V}....(i)and(8−37)20=Wand (8-3 \sqrt{7})^{20}=\mathrm{W}....(ii) Here, o<W<1\mathrm{o} < \mathrm{W} < 1 Now, adding equation (i) \& (ii), we have, U+V+W=(8+37)20+(8−37)20=2[20C0⋅820+20C2818⋅(37)2+….+(37)20]\begin{aligned} & \mathrm{U}+\mathrm{V}+\mathrm{W}=(8+3 \sqrt{7})^{20}+(8-3 \sqrt{7})^{20} \\ & =2\left[{ }^{20} \mathrm{C}_0 \cdot 8^{20}+{ }^{20} \mathrm{C}_2 8^{18} \cdot(3 \sqrt{7})^2+\ldots .+(3 \sqrt{7})^{20}\right] \end{aligned} R.H.S. is an even number. Also, o<V<1,o<W<1\mathrm{o} < \mathrm{V} < 1, \mathrm{o} < \mathrm{W} < 1 and U is an integer ⇒V+W\Rightarrow \mathrm{V}+\mathrm{W} is an integer ⇒V+W=1\Rightarrow \mathrm{V}+\mathrm{W}=1(U+V)W=(8+37)20(8−37)20=[(8+37)(8−37)]20=(64−63)20=120=1\begin{aligned} & (U+V) W=(8+3 \sqrt{7})^{20}(8-3 \sqrt{7})^{20} \\ = & {[(8+3 \sqrt{7})(8-3 \sqrt{7})]^{20} } \\ = & (64-63)^{20}=1^{20}=1\end{aligned}$$
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