Question 4 (b)

Find the sum of:
–52 and 52

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Solution

(–52) + 52 = -52 + 52 = 0

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- 52

52

find the sum of 52^{\circ} 38^{'} and 37^{\circ} 42^{'} 30^{''} .

If the sum of the first ten terms of the series

${(1 \frac{3}{5})}^{2} + {(2 \frac{2}{5})}^{2} + {(3 \frac{1}{5})}^{2} + 4^{2} + {(4 \frac{4}{5})}^{2} + . . . . .,$

Is $\frac{16}{5}$ m, then m is equal to:

^{47} C_{4} + 5 \sum r = 1

^{52 - r} C_{3} = ?

1^{2} - \frac{2^{2}}{5} + \frac{3^{2}}{5^{2}} - \frac{4^{2}}{5^{3}} + \frac{5^{2}}{5^{4}} - \frac{6^{2}}{5^{5}} + . . . .

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