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Let `a=3^(1/(223))+1`and for all `geq3,l e tf(n)=^n C_0dota^(n-1)-^n C_1dota^(n-2)+^n C_2dota^(n-3)-+(-1)^(n-1)dot^n C_(n-1)dota^0`. If the value of `f(2007)+f(2008)=3^k w h e r ek in N ,`then the value of `k`is. |
Answer» Correct Answer - 9 `f(n)=.^(n)C_(0)a^(n-1)-.^(n)C_(1)a^(n-2)+.^(n)C_(2)a^(n-3)+"...."+(-1)^(n-1).^(n)C_(n-1)a^(0)` `= 1/a(.^(n)C_(0)a^(n)-.^(n)C_(1)a^(n-1)+.^(n)C_(2)a^(n-2)+"...."+(-1)^(n-1)C_(n-1)a)` `=1/a((a-1)^(n)-(-1)^(n).^(n)C_(n))` `= 1/a((3^(n/223)-(-1)^(n)))` `f(x) = (3^(n/223)-(-1)^(n))/((3^(1/223)+1))` `rArr f(2007)= (3^(2007/223)+1)/(3^(1/223)+1)` `rArr f(2008)= (3^(2008/223)+1)/(3^(1/223)+1)` `f(2007)+f(2008)= (3^(2007/223)+3^(2008/223))/(3^(1/223+1))` `= (3^(9)+3^(9+(1)/(223)))/(3^(1/223)+1)` `= 3^(9)((1+3^(1/223)))/((1+3^(1/223)))=3^(9)` `rArr 3^(9)= 3^(k)` then `k = 9` |
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