Prove that 2n^3 + 3n^2 + n is divisible by 6 for every integer n > 1 by mathematical induction. (7 marks)

The answer is True

Step-by-step explanation:

A mathematical induction consists in only 2 steps:  

First step: Show the proposition is true for the first one valid integer number.

Second step: Show that if any one is true then the next one is true  

Finally, if first step and second step are true, then the complete proposition is true.

So, given

First step: using and replacing n=2 (the first valid integer number >1)

As the result is an integer number, so the first step is true.

Second step: using any next number, , let it replace

As the First step is true, we know that

,

So let it replace in the previous expression

Finally

where the last expression is an integer number

So the second step is true, and the complete proposition is True

exactly you go it

step-by-step explanation:

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)