Question

calculate 2+4+6+8+2n for n=1 and 2

Answer 1

Answer:

sry I can't but I can do you want some

Answer 2

Answer:

There is a proof method called “math induction”. It consists of 2 steps.

Check that it is true for n=1. Really, 2 = 1(1+1)

Now let’s assume that there is such k, that for every n<=k the equality takes place. And let’s prove that the equality takes place for n = k+1.

So, we assume that we know that 2+4+6+8+…+2k = k(k+1)

Let’s see, what will be equal to 2+4+6+8+…+2k+2(k+1)

All the added numbers except the last one can be reduced into the product:

2+4+6+8+…+2k+2(k+1) = k(k+1)+2(k+1)

Now we extract the common multiplier:

k(k+1)+2(k+1) = (k+2)(k+1) = (k+1)((k+1)+1)

So, really,

2+4+6+8+…+2k+2(k+1) = (k+1)((k+1)+1)

So, in other world: we check that the equality is true for n=1, and then we prove that it’s possible to make a “step of induction” and increase the number by 1, and the equality will still exist. This proves that the equality will be true for any n = 2, 3, 4, 5, 6,…till infinity.

Hope it helps you :D

#StaySafe

#KeepLearning