Mathematical Induction getting the right sideSo I 've been doing Mathematical Inductions but I seem to have a issue in simplify and getting the right side.So I have this on the L.H.S k ( k + 1 ) ( 2 k + 1 ) 6 + ( k + 1 ) 2 And I'm trying to make it equal to ( k + 1 ) ( k + 2 ) ( 2 k + 3 ) 6 I've check the book for help but for this it seems to be skipping some steps.And I might as well post this as well which I had the same issue as above. 3 ( 5 k + 1 − 1 ) 4 + 3 · 5 k + 1 to 3 ( 5 k + 2 − 1 ) 4 .

Question

Mathematical Induction getting the right sideSo I &#039;ve been doing Mathematical Inductions but I seem to have a issue in simplify and getting the right side.So I have this on the L.H.S            k      (      k      +      1      )      (      2      k      +      1      )        6    +  (  k  +  1      )    2  And I&#039;m trying to make it equal to            (      k      +      1      )      (      k      +      2      )      (      2      k      +      3      )        6  I&#039;ve check the book for help but for this it seems to be skipping some steps.And I might as well post this as well which I had the same issue as above.            3      (              5                  k          +          1                    −      1      )        4    +  3  ·      5          k      +      1      to            3      (              5                  k          +          2                    −      1      )        4    .

Oliver Shepherd · Accepted Answer

(1)            k      (      k      +      1      )      (      2      k      +      1      )        6    +  (  k  +  1      )    2    =            k      (      k      +      1      )      (      2      k      +      1      )      +      6      (      k      +      1              )        2              6    =  (  k  +  1  )            k      (      2      k      +      1      )      +      6      (      k      +      1      )        6    =  (  k  +  1  )            k      (      2      k      +      1      )      +      6      k      +      6        6    =  (  k  +  1  )            2              k        2            +      7      k      +      6        6    =            (      k      +      1      )      (      k      +      2      )      (      2      k      +      3      )        6  (2) Is your second statement even true? for   k  =  1  I get:   34  =  93Let       5    k    =  x            3      (      5      x      −      1      )        4    +  3  x  +  1  =            15      x      −      3      +      12      x      +      4        4    =            27      x      +      1        4  which doesn&#039;t seem to be equal to your given RHS

Gretchen Schwartz · Accepted Answer

k      (      k      +      1      )      (      k      +      2      )        6    +  (  k  +  1      )    2  Let us get the part of this expression on the right side to have a common denominator with the fraction, let us use 6. We now have:            k      (      k      +      1      )      (      2      k      +      1      )        6    +            6      (      k      +      1              )        2              6    =            k      (      k      +      1      )      (      2      k      +      1      )      +      6      (      k      +      1              )        2              6  Factoring out a   (  k  +  1  ) from the numerator:  =            (      k      +      1      )      (      k      (      2      k      +      1      )      +      6      (      k      +      1      )      )        6  Expanding the expression inside the parentheses:  =            (      k      +      1      )      (      2              k        2            +      k      +      6      k      +      6      )        6    =            (      k      +      1      )      (      2              k        2            +      7      k      +      6      )        6  Now we can factor   2      k    2    +  7  k  +  6 into   (  k  +  2  )  (  2  k  +  3  )  and we obtain: =            (      k      +      1      )      (      k      +      2      )      (      2      k      +      3      )        6