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Let be $a_1$, $a_2$, $\ldots$, $a_{n-1}$, $a_n$, term of the arithmetics sequence. Please help me prove that


is sum the n-th term of the given sequence. Thanks.

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HINT: $a_1+a_n=a_2+a_{n-1}=\ldots$ –  Raskolnikov Oct 10 '12 at 7:57
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2 Answers

up vote 0 down vote accepted

Let be $1, 2, 3, 4, 5, 6, 7, 8, 9, 10$ real numerical sequence. That one sequence let be arithmetics sequence must that


Note that the given sequence is


therefore the given sequence is arithmetics sequence. Also note that


So generally worth

$(*)$ $\ldots$ $a_1+a_n=a_2+a_{n-1}=a_3+a_{n-2}=\cdots$

Now turn prove the given formula.

The given sequence is

$a_1, a_2, \ldots, a_{n-1}, a_n$.

Form the sum of the given sequence, and that sum the mark $S_n$, i.e

$(1)$ $\ldots$ $S_n=a_1+a_2+\ldots+a_{n-1}+a_n$

Because comutative low for the sum (+) is worth, have

$(2)$ $\ldots$ $S_n=a_n+a_{n-1}+\ldots+a_2+a_1$

Equation $(1)$, $(2)$ collect side to side



Implement the $(*)$ we have:




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Thanky sir, for proving –  user39471 Oct 10 '12 at 8:11
How did you manage to type this up in less than 3 minutes? –  commenter Oct 10 '12 at 8:21
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Writing the arithmetic sequence in another way, $$a_0 + (a_0 + d) + (a_0 + 2d) + ...$$

Flip it around to get another sequence that decreases to $a_0$,

$$ (a_0 + nd) + (a_0 + (n-1)d) + ...$$

Notice that the sum of the $k$-th term of one sequence with the other is always $(2a_0 + nd) = (a_0 + a_n)$.

There are $n$ such terms. So $n(a_0 + a_n)$ is twice the sum of the sequence.

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