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Show that the integer next above ({3} , + , 1)^{2m} contains

Question

Show that the integer next above $(3 + 1)^{2 m}$ contains $2^{m + 1}$ as a factor

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Sebastian Discussed

Show that the integer next above

(3 + 1)^{2 m}

contains

2^{m + 1}

as a factor

7 mins ago

Discuss this question LIVE

7 mins ago

Text solutionVerified

Let

(3 + 1)^{2 m}

= I + f
and

(3 - 1)^{2 m}

= f' where 0 < f' < 1
and 0 < f' < 1 and I is an integer.
Thus I + f + f' =

(3 + 1)^{2 m} + (3 - 1)^{2 m}

(4 + 23)^{m} + (4 - 23)^{m}

2^{m} [(2 + 3)^{m} + (2 - 3)^{m}]

2^{m + 1} [2^{m} +^{m} C_{2} . 2^{m - 2} . (3)^{2} + .....]

Now as in part (a) f + f' = 1 and so I + f + f' is an integer next above

(3 + 1)^{2 m}

which by (1) contains

2^{m + 1}

as a factor.

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Question Text	Show that the integer next above $(3 + 1)^{2 m}$ contains $2^{m + 1}$ as a factor
Answer Type	Text solution:1
Upvotes	150