Question

For any integer m>=3, the series 2+4+6+…+(4m) can be equivalent to

Answer 1

we need to find the sum of ..

2+4+6+...+4m

Explanation:

first 2 can be taken common from the given sequence

2(1+2+3+...+2m)

then it becomes..

2(2m)(2m+1)/2..two will be removed..

therefore it becomes..

2m(2m+1) ..

minimum value of this sequence is

2×3(2×3+1)

=42 it's minimum value at m=3

HOPE THIS HELPS YOU..

Answer 2

Given :   2+4+6+…+(4m)  

To find : Equivalent

Solution:

2 + 4 + 6  + .......................................................+ 4m

= 2(1  + 2  + 3 + .............. + 2m)

= 2 ( 2m)(2m + 1)/2

= 2m(2m + 1)

for m = 3

= 2(3)(2*3 + 1)

= 6 * 7

= 42

Series is Equivalent to  2m(2m + 1) ≥  42  for m ≥ 3

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