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Let 10∑k=1f(a+k)=16(210−1), where the function f satisfies f(x+y)=f(x)+f(y) for all natural numbers x,y and f(1)=2. Then the natural number 'a' is: |
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Answer» Let 10∑k=1f(a+k)=16(210−1), where the function f satisfies f(x+y)=f(x)+f(y) for all natural numbers x,y and f(1)=2. Then the natural number 'a' is: |
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