Use mathematical induction to prove that each of the following is true for all natural numbers n.

Question

2 + 4 + 6 + ... + (2n) = n(n+1)

Answer 1

for n=1

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

for n=2

2+4 = 2(2+1) = 6

assume for n=k

2 + 4 + 6 ...2k = k(k+1)

try to prove for n=k+1

2 + 4 + 6 +..+ 2k + 2(k+1) = (k+1)(k+1+1)

assume 2 + 4 + 6 +2k = k(k+1)

add 2(k+1) to both sides

2+ 4 + 6 +... + 2k + 2(k+1) = k(k+1) + 2(k+1) = (k+1)(k+2) = (k+1)(k+1 +1)

2+4 + 6 +... + 2k + 2(k+1) = (k+1)(k+1 +1)

QED

Answer 2

This is the sum of an arithmetic sequence starting the first element a₁=2, a₂=4, ... a_n=2n

Arithmetic sequence is defined as a list of sequential numbers where the the difference between one term and next term is a constant value.

{1,2,3,4,5,6...,n} is considered a basic arithmetic sequence that shall be remembered.

The sum of the above list is (1+n)*n/2

If you could not remember it, it can be inducted in the following way.

If n is an even number, like 2m (m≥1)

then try to combine the first element with the last element, i.e, 1 + 2m

then combine the second element with the last but one element, i.e, 2 + (2m-1) = 2m +1

then combine the third element with the last but two element, i.e, 3+(2m-2)=2m+1.

...

and so on, until there is no more combination. Since n is even number, there will be total of m combinations.

So the sum of the list is m(2m+1) , substitute m=n/2 into m(2m+1), the sum is now n(n+1)/2;

The same induction can be done if n is an odd number by using n = (2m+1) m≥1

The list {2, 4, 6, ..., 2n} is similar to the basic sequence {1,2,3,..., n}, the only difference is each term is twice of the basic sequence term. The above arithmetic sequence summation can be written in a general formula like (a₁+a_n)×n/2

n

∑ a_i = (a₁+a_n) × n/2 = (2+2n)×n/2 =(n+1)n

i=1

Answer 3

a. Demonstrate true for n = 1

2(1) (n = 1(1 + 1)

b. Assume true for n

2 + 4 + 6 + ... + 2n = n(n + 1)

c. Demonstrate true for n + 1

2 + 4 + 6 + ... + 2n + 2(n + 1) = n(n + 1) + 2(n + 1)

2 + 4 + 6 + ... + 2n + 2(n + 1) = (n + 1)(n + 2)

2 + 4 + 6 + ... + 2n + 2(n + 1) = (n + 1)(n + 1 + 1)

QED