Question 143(i)

Use the propeties of exponents to verify that each statement is true.
$\frac{1}{4} (2^{n}) = 2^{n - 2}$

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Solution

True:
$\frac{1}{4} (2^{n}) = 2^{n - 2} R H S = 2^{n - 2} = 2^{n} \div 2^{2} [∵ a^{m} \div a^{n} = (a)^{m - n}] = \frac{2^{n}}{4} = L H S$

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Question 143(i)

Use the propeties of exponents to verify that each statement is true.
$\frac{1}{4} (2^{n}) = 2^{n - 2}$

Question 143(iii)

Use the propeties of exponents to verify that each statement is true.
$25 (5^{n - 2}) = 5^{n}$

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P (n)

" 3 n + 1

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is true but

P (2)

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Verify that for all $n \geq 1$ , the sum of the squares of the first 2n positive integers is given by the formula $1^{2} + 2^{2} + 3^{2} + . . . . . (2 n)^{2} = \frac{n (2 n + 1) (4 n + 1)}{3}$

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Question 143(i) Use the propeties of exponents to verify that each statement is true. 14(2n)=2n−2

True: 14(2n)=2n−2RHS=2n−2=2n÷22 [∵ am÷an=(a)m−n]=2n4=LHS

Question 143(i)

Use the propeties of exponents to verify that each statement is true.
$\frac{1}{4} (2^{n}) = 2^{n - 2}$

True:
$\frac{1}{4} (2^{n}) = 2^{n - 2} R H S = 2^{n - 2} = 2^{n} \div 2^{2} [∵ a^{m} \div a^{n} = (a)^{m - n}] = \frac{2^{n}}{4} = L H S$