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Without adding, find the sum (1+3+5+7+9+11+13+15+17+19). |
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Answer» Given (1+3+5+7+9+11+13+15+17+19) According to the property of perfect square, for any natural number n, we have sum of first n odd natural numbers = n2 But here n = 10 Applying the above law we get Therefore (1+3+5+7+9+11+13+15+17+19) = 102 = 100 |
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