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if a,b,c,d,e and f are six real numbers such that a+b+c=d+e+fa^2+b^2+c^2=d^2+e^2+f^2 and a^3+b^3+c^3=d^3+e^3+f^3 , prove by mathematical induction that a^n+b^n+c^n=d^n+e^n+f^n forall n in N. |
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Answer» Solution :Let `P(n):a^(n)+b^(2)+c^(n)=d^(n)+e^(n)+f^(n),AA n in N ""..(i)` where `a+b+c+d=e+""...(ii)` `a^(2)+b^(2)+c=d^(2)+e^(2)+f ""....(III)` and `a^(2)+b^(3)+c^(3)=d^(3)+e^(2)+^(3)""...(IV)` Step I from n from Eq. (i) we get `P(1): a+b+c=d+e+f"" ` [ given] Hence the result is true for n 1 Also, for n=2 from Eq(i), we get `P(2): a^(2)+b^(2)+c^(2)=d^(2)+e^(3)+f^(3) ""` [ given] Hennce the result trueor n=3 Therefore, P(1) , P(2) and P (3) are true. Step II Assume that `P(k-2),P(k-1)and P(k)` are true, then `P(k-2), a^(k-2)+b^(k-2)= d^(k-2)+e^(k-2)+f^)k-2) ""...(v)` `p(k-1):a^(-1)+b^(k-1)+c^(k-1)=d^(k-1)+e^(k-1)+f^(k-1) ""....(vi)` and `P(k): a^(k)+b^(k)+c^(k)=d^(k)+e^(k)+f^(k) "m"...(vii)` Step III for ` xn=k+1` we SHALL to prove that `P(k+1):a^(k+1)+b^(k+1)=d^(k+1)+e^(k+1)+f^(k+1)` LHS `=a^(k+1)+b^(k+1)+c^(k+1)` `=(a^(k)+b^(k)(a+b+c)-(a^(k-1)+b^(k-1)+c^(k-1))` `(ab+bc+ca)+abc(a^(k-2)+b^(k-2)+b^(k-2)+c^(k-2))` `=(d^(k)+e^(f)+f^(k))(d+e+f)-(d^(k-1)+e^(k-1)+c^(k-2))` `(de+ef+fd)+def(d^(k-2)+e^(k-2)+f^(k-2))` [ using Eqs. (ii), (iii), (iv), (v), (vi), (vii)] `:. (a+b+c)^(2)=(d+e+f)^(2)` `RARR a^(2)+b^(2)+c^(2)+2(ab+bc+ca)` `=d^(2)+e^(2)+f^(2)+2(de+ef+fd)` `rArr ab+bc+ca=de+ef+fd` `[ :. a^(2)+b^(2)+c^(2)=d^(2)+e^(2)+f^(2)]` and `a^(3)+b^(3)+c^(3) -3abc` `(=d+e+f)(d^(2)+e^(2)+f^(2)-de-ef-fd)` `=d^(3)+e^(3)+f^(3)-EDF` `rArr abc=def [ :. a^(3)+b^(3)+c^(3)=d^(3)+e^(3)+f^(3))` `=d^(k+1)+e^(k+1)+f^(k-1)=RHS` This shows that result is true for n=k+1. Hence by second perincipal of mathmatical inducition, the result is true for all `n in N`. |
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