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Chapter 8: Problem 604

If \((666 \ldots \text { n times })^{2}+(8888 \ldots\) n times \()=(4444 \ldots \mathrm{K}\) times \()\) then \(\mathrm{K}=\) (A) \(n+1\) (B) \(\mathrm{n}\) (C) \(2 \mathrm{n}\) (D) \(\mathrm{n}^{2}\)

Short Answer

Expert verified
The short answer is: To find the value of K, compare the simplified sum's expression of \((6^2n + 8^2n)(10^{2(n-1)} + 10^{2(n-2)} + \cdots + 10^0)\) with the given result in terms of 4: \(4(10^{2(n-1)} + 10^{2(n-2)} + \cdots + 10^0)\). After simplifying, we find that \(K = n\), which corresponds to answer choice \(\boxed{\text{(B) }n}\).

Step by step solution

01

Square the given numbers

We have two numbers: 666... (n times) and 888... (n times). First, we need to square these numbers. The square of 666... (n times) is: \(((6\times10^{(n-1)} + 6\times10^{(n-2)} + \cdots + 6\times10^0))^2\), The square of 888...(n times) is: \(((8\times10^{(n-1)} + 8\times10^{(n-2)} + \cdots + 8\times10^0))^2\).
02

Add the squared numbers

Now, we need to add the squares of these numbers to get the sum: \(((6\times10^{(n-1)} + 6\times10^{(n-2)} + \cdots + 6\times10^0))^2 + ((8\times10^{(n-1)} + 8\times10^{(n-2)} + \cdots + 8\times10^0))^2\).
03

Simplify the expression

Let's simplify the given expression: \((6^2 + 6^2 + \cdots + 6^2)(10^{2(n-1)} + 10^{2(n-2)} + \cdots + 10^0) + (8^2 + 8^2 + \cdots + 8^2)(10^{2(n-1)} + 10^{2(n-2)} + \cdots + 10^0)\). Factor out the common term, (10^{2(n-1)} + 10^{2(n-2)} + \cdots + 10^0): \((6^2 + 6^2 + \cdots + 6^2 + 8^2 + 8^2 + \cdots + 8^2)(10^{2(n-1)} + 10^{2(n-2)} + \cdots + 10^0)\). Combine the like terms: \((6^2n + 8^2n)(10^{2(n-1)} + 10^{2(n-2)} + \cdots + 10^0)\).
04

Define the result in terms of K

Since the result is given as 444...(K times), let's represent it with the number 4: \(4(10^{2(n-1)} + 10^{2(n-2)} + \cdots + 10^0)\).
05

Find the value of K

Comparing the simplified sum's expression with the given result in terms of 4: \((6^2n + 8^2n)(10^{2(n-1)} + 10^{2(n-2)} + \cdots + 10^0) = 4(10^{2(n-1)} + 10^{2(n-2)} + \cdots + 10^0)\). Divide both sides by (10^{2(n-1)} + 10^{2(n-2)} + \cdots + 10^0): \(6^2n + 8^2n = 4\). Solve for K: \(K = n\). The correct answer is (B) n.

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Most popular questions from this chapter

If the angles \(\mathrm{A}<\mathrm{B}<\mathrm{C}\) of a \(\triangle \mathrm{ABC}\) are in \(\mathrm{A} . \mathrm{P}\). then (A) \(c^{2}=a^{2}+b^{2}-a b\) (B) \(c^{2}=a^{2}+b^{2}\) (C) \(b^{2}=a^{2}+c^{2}-a c\) (D) \(a^{2}=b^{2}+c^{2}-b c\)

If \([(b+c-a) / a],[(c+a-b) / b],[(a+b-c) / c]\) are in A. \(P\) then \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are in (A) G P. (B) A. P. (C) H. P. (D) A. G P.

Sum to infinity of the series \(1+(4 / 5)+\left(7 / 5^{2}\right)+\left(10 / 5^{3}\right)+\ldots\) is (A) \((5 / 16)\) (B) \((35 / 16)\) (C) \((16 / 35)\) (D) \((7 / 16)\)

If the first, second and last terms of an A. P. are a, b and \(3 a\) respectively, the sum of the series is (A) \(\left[\left(4 a^{2}\right) /(b-a)\right]\) (B) \(\left[\left(2 a^{2}+2 a b\right) /(b-a)\right]\) (C) \(\left[\left(2 a b+a^{2}\right) /(b-a)\right]\) (D) \(\left[\left(2 a^{2}-2 a b\right) /(a-b)\right]\)

\(2+12+36+80+\ldots \mathrm{n}\) terms \(=\) (A) \([\\{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)(3 \mathrm{n}+5) \overline{\\} /(24})]\) (B) \([\\{n(n+1)(n+2)(3 n+1)\\} /(12)]\) (C) \([\\{n(n+1)(n+3)(n+5)\\} /(24)]\) (D) \([\\{n(n+1)(n+2)(n+3)\\} /(12)]\)
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