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The sum 1/(1!9!) + 1/(3!7!) + ... + 1/(7!3!) + 1/(9!1!) can be written in the form 2^a/(b!) Find a and b. |
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Answer» Solution :1/(1!9!) + 1/(3!7!) + ... + 1/(7!3!) + 1/(9!1!)` `1/(10!)((10!)/(1!9!) + (10!)/(3!7!) + .... + (10!)/(9!1!))` `1/(10!) ("^10C_1 + ^10C_3 + .... + ^10C_9)` `1/(10!) 2^(10-1)` `2^9/(10!)` = `2^a/(B!)` where a = 9, b = 10 |
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