Using the formula for squaring a binomial, evaluate the following:

$(i) (102)^{2}$ $(i i) (99)^{2}$ $(i i i) (1001)^{2}$ $(i v) (999)^{2}$ $(v) (703)^{2}$

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Solution

$(i) (102)^{2} = (100 + 2)^{2} = (100)^{2} + 2 \times 100 \times 2 + (2)^{2} {(a + b)^{2} = a^{2} + 2 a b + b^{2}} = 10000 + 400 + 4 = 10404 (i i) (99)^{2} = (100 - 1)^{2} = (100)^{2} - 2 \times 100 \times 1 + (1)^{2} {(a - b)^{2} = a^{2} - 2 a b + b^{2}} = 10000 - 200 + 1 = 10001 - 200 = 9801 (i i i) (1001)^{2} = (1000 + 1)^{2} {(a + b)^{2} = a^{2} + 2 a b + b^{2}} = (1000)^{2} + 2 \times 1000 \times 1 + (1)^{2} = 1000000 + 2000 + 1 = 1002001 (i v) (999)^{2} = (1000 - 1)^{2} {(a - b)^{2} = a^{2} - 2 a b + b^{2}} = (1000)^{2} - 2 \times 1000 \times 1 + (1)^{2} = 1000000 - 2000 + 1 = 1000001 - 2000 = 998001 (v) (703)^{2} = (700 + 3)^{2} = (700)^{2} + 2 \times 700 \times 3 + (3)^{2} {(a + b)^{2} = a^{2} + 2 a b + b^{2}} = 490000 + 4200 + 9 = 494209$

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(101)^{2}

Using the formula for squaring a binominal evaluate the following:
(i) $(54)^{2}$
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(iii) $(103)^{2}$
(iv) $(704)^{2}$

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