what is mathematical induction of 2+4+6+.......+2n=n^2+n?

Question

plz do it step by step so i can understand

Answer 1

Induction is divided into 3 parts.

First, prove it, when n=1
When n=1,
2*1 = 1^2+1
2=1+1
2=2
For n=1, the formula was right.

Second, prove it like this :
If n=k was right, then
n=k+1 was right too...
For n=k,
2+4+6+...+2k=k^2+k --> This is 1st Equation
For n=k+1,
2+4+6+...+2(k+1) = (k+1)^2+(k+1)
2+4+6+...+2k+(2k+2) = k^2+2k+1+k+1
[2+4+6+...+2k] + 2k + 2 = k^2+3k+2
Subsitute 1st Equation into the equation above, and you get:
k^2 + k + 2k + 2 = k^2+3k+2
k^2+3k+2=k^2+3k+2
So, when n=k was right, n=k+1 was right too...

Third, prove it for all condition :
If n=1 was right (from first statement), then n=2 was right too (from 2nd statement)
If n=2 was right, n=3 was right too...
And so, n is right for all natural numbers.

You can apply induction only on natural numbers provided, just like in this cases, when n is natural number.

Answer 2

2 + 4 + 6 + ... + 2n = n^2 + n

Proof (by induction).

1) Base Case: Let n = 1. Then
LHS = 2 ... 2 = 2
RHS = 1^2 + 1 = 1 + 1 = 2
LHS = RHS, so the formula holds true for n = 1.

2) Induction hypothesis: Assume the formula holds true for n = k. That is, assume that

2 + 4 + 6 + ... + 2k = k^2 + k for some value k.

(We want to prove that
2 + 4 + 6 + ... + 2(k + 1) = (k + 1)^2 + (k + 1) )

BUT the sum of (k + 1) terms is the same as the sum of k terms plus the (k + 1)th term. i.e.

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

And, as you can see, within the square brackets is our induction hypothesis. Replace with k^2 + k.

2 + 4 + 6 + ... + 2(k + 1) = [ k^2 + k ] + 2(k + 1)

Expand the right hand side.

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

And manipulate to what we know the answer is going to be.

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

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

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

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

Showing that the formula holds true for n = k + 1.

Thus by the principle of mathematical induction, the formula holds true for all natural numbers n.

Answer 3

Let P(k) be proposition that :-
2 + 4 + 6 + -------2k = k² + k
Have now to show P(1) true and P(k + 1) true.

Consider P(1):-
LHS = 2
RHS = 1² + 1 = 2
Thus P(1) is true
Consider P(k + 1):-
2 + 4 + 6 ------2k + 2(k + 1) = (k + 1)² + (k + 1)
Have now to show that P(k + 1) is true:-
2 + 4 + 6 + -----2k = k² + k (is true)
2 + 4 + 6 ---2k + (2k + 2) = k² + k + (2k + 2)
2 + 4 + 6 --2k + 2(k + 1) = (k² + 2k + 1) + (k + 1)
2 + 4 + 6--2k + 2(k + 1) = (k + 1)² + (k + 1)
Thus P(k + 1) is true.
Conclusion
P(k) true => P(1) true , P(k + 1) true
Have therefore proved by induction that P(n) is true as stated in question.

Answer 4

for n =1

LHS = 2 and RHS = n^2+n = 1+1 = 2

proved for init step
now step for induction

for n let it be true

add 2(n+1)
RHS = n^2+n +2(n+1)
= n^2 + 2n + 1 + n+1
= (n+1)^2+(n+1)

if f(n) = n^2 +n then
f(n+1) = (n+1)^2 + (n+1)

so LHS = RHS

induction step is proved

hence proved

Answer 5

2+4+6+.......+2n=n^2+n

Check if it is true for n=1
2=1^2+1=2 True

Assume it is true for n=m
2+4+6+...+2m=m^2+m

Add 2(m+1) to both sides
2+4+6+...+2m+2(m+1)
=m^2+m+2(m+1)

RHS =
m^2+m+2m+2
=(m^2+2m+1)+m+1
=(m+1)^2+(m+1)

If it is true for n=m, the it will be true for n=m+1

It is true for n=1
Must be true for n=1+1 .....
True for all values of n

Answer 6

First of all please forgive my English gramma because i'm a taiwanese
2+4+6+...+2n--------A
reverse
2n+...+6+4+2--------B
A+B=(2+2n)+[4+(2n-2)]+[6+(2n-4)]+...+(2n+2)
=(2+2n)+(2+2n)+(2+2n)+...+(2+2n)
=(2+2n)n------because there are n items to add------note
A=(A+B)/2=(2+2n)n/2
=(2n+2n^2)/2=n+n^2----->your answer n^2+n

hope you can understand the "note"
for example:
1+2+3+...+10 there are 10 items to add
1+2+3+...+n there are n items to add
so 2+4+6+...+2n there are n items to add