Mathematical Induction getting the right sideSo I 've been doing Mathematical Inductions but I seem...

grenivkah3z

grenivkah3z

Answered

2022-07-04

Mathematical Induction getting the right side
So 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 .

Answer & Explanation

Oliver Shepherd

Oliver Shepherd

Expert

2022-07-05Added 24 answers

(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 = 93
Let 5 k = x
3 ( 5 x 1 ) 4 + 3 x + 1 = 15 x 3 + 12 x + 4 4 = 27 x + 1 4
which doesn't seem to be equal to your given RHS
Gretchen Schwartz

Gretchen Schwartz

Expert

2022-07-06Added 5 answers

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

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