Proving by induction. We'd like to show that $2 + 4 + 6 + \cdots+ 2n = n(n + 1)$.
A nice way to do this is by induction. Let $S(n)$ be the statement above. An inductive proof would have the following steps: Show that $S(1)$ is true. Show that if $S(1),\ldots,S(k)$ are true, then so is $S(k + 1)$.
This question is really starting to bug me, am I doing something wrong or is the equation wrong for that series? It only seems to work for $S(1)$ but after that it does not give the correct series. The $n(n+1)$ series is $2, 6, 12,\ldots.$ I'm really confused can someone please nudge me in the right direction (I know how proofs by induction works) I'm just having problems with this one in particular.
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$\begingroup$First, that answer is correct. If you note that $2+4+6+...+2n = 2*(1+2+3+...+n)=2 \frac{n*(n+1)}2 = n*(n+1)$
Second. Suppose that it is true, you have $2+4+6+...+2n = n*(n+1)$ This is $S(n)$ add 2(n+1) to both sides, giving $$2+4+...+2n+2(n+1) = n*(n+1) + 2(n+1) = (n+2)(n+1)$$ which is $S(n+1)$
$\endgroup$ $\begingroup$If $S(k)$ is true, we have $\sum_{i=1}^k2i=k(k+1)$ Then we want to evaluate $\sum_{i=1}^{k+1}2i=\sum_{i=1}^k2i+2(k+1)=k(k+1)+2(k+1)=(k+2)(k+1)$ so $S(k+1)$ is true.
$\endgroup$ $\begingroup$Base case: $n=1$
$2=1(1+1)=2$
Assume true for $n=k$, that is:
$2+4+6+...+2k=k(k+1)$
To show true for $n=k+1$
$2+4+6+...+2k+2(k+1)=k(k+1)+2(k+1)=(k+1)(k+2)=(k+1)(k+1+1)$
Hence, $2+4+6+...+2n=n(n+1)$
The equation is correct. It is just a simple arithmetic series. If you can remember a formula for an arithmetic series given by $S_n=\frac{n(a_1+a_n)}{2}$. In this case the first term is $2$ the last term is $2n$ and the number of terms is $n$ so we have:
$$S_n=\frac{n(2+2n)}{2}=\frac{2n(1+n)}{2}=n(n+1)$$
$\endgroup$ $\begingroup$I'll try to explain this in plain English.
Problem statement: 2+4+6+...+2n = n(n+1)
The statement can be broken down into 3 parts.
a) The portion on the left-hand side of the = sign, before the ... i.e. 2+4+6, which represents a sequence of numbers.
b) The portion of the left-hand side of the = sign, after the ... i.e. 2n, which is used to calculate the element of the sequence at the index n. Note the sequence is 1 based, not 0 based.
c) Everything on the right-hand side of the = sign, i.e. n(n+1), which is used to sum up all elements within the sequence.
Consider the following when n=5
2n = 2*5 = 10, therefore the sequence can be written as 2+4+6+?+10.
Note the 4th element of the sequence is currently unknown, which isn't an impediment, as it can be resolved later using elementary arithmetic.
Use the formula on the right-hand side of the = sign, to sum together all elements within the sequence, including the unknown values as follows: n(n+1) = 5(5+1) = 5*6 = 30.
Then use basic algebra to find the missing element at index 4: 2+4+6+?+10 = 30 22+?=30 ?=30-22 ?=8
Therefore, the resulting sequence is 2+4+6+8+10
Hope this helps :-)
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