Sets

Sets

Question

Prove the following by using the principle of mathematical induction for all straight n space element of space straight N.

41 to the power of straight n space minus space 14 to the power of straight n is a multiple of 27 for all straight n space element of space straight N.



Answer

Let P(n): 41 to the power of straight n space minus space 14 to the power of straight n is a multiple of 27.
I.       For n = 1,
        P(1): 41 to the power of 1 space minus space 14 to the power of 1 is a multiple of 27
rightwards double arrow    41 - 14 is a multiple of 27 rightwards double arrow 27 is a multiple of 27
         which is true.

∴       P(n) is true for n = 1
II.    Suppose P(n) is true for n = m, straight m space element of space straight N.
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#6 {main}</pre>  ...(i)
III.   For n = m + 1,
        P (m + 1) : 41 to the power of straight m plus 1 end exponent space minus space 14 to the power of straight m plus 1 end exponent is a multiple of 27
        But, 41 to the power of straight m plus 1 end exponent minus space 1 space 4 to the power of straight m plus 1 end exponent space equals space 41 to the power of straight m plus 1 end exponent space minus space 41 to the power of straight m. end exponent 14 space plus space 41 to the power of straight m.14 space minus space 14 to the power of straight m plus 1 end exponent
           
           = 41 to the power of straight m left parenthesis 41 minus 14 right parenthesis space plus 14 space left parenthesis 41 to the power of straight m minus 14 to the power of straight m right parenthesis space equals space 41 to the power of straight m space cross times space 27 space plus space 14 space left parenthesis 27 straight k right parenthesis       [By (i)]
           equals 27 space left square bracket 41 to the power of straight m space plus space 14 straight k right square bracket space equals space 27 straight k apostrophe  where straight k apostrophe space equals space 41 to the power of straight m space plus space 14 space straight k space element of space straight Z
rightwards double arrow space space space space space space space space space space 41 to the power of straight m plus 1 end exponent space minus space 14 to the power of straight m plus 1 end exponent is a multiple of 27

∴           P(m + 1) is true.

∴           P(m) is true rightwards double arrowP(m + 1) is true.
Hence, by induction, P(n) is true for all straight n space element of space straight N.


      

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