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Using binomial theorem, find the value of `(0.99)^(15)` up to four places of decimal. |
Answer» We have `(0.99)^(15)=(1-0.01)^(15)` `=1-.^(15)C_(1) xx(0.01) + .^(15)C_(2) xx (0.01)^(2)-.^(15)C_(3) xx (0.01)^(3) +...` [neglecting higher powers of 0.01] `=1-15 xx(0.01) + 105 xx (0.0001)-455 xx (0.000001)+...` `=1-0.15+0.0105 + 0.000455 = 0.860045.` Hence, `(0.99)^(15) = 0.860045.` |
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