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$(a_1,a_2,a_3,..)$ be a sequence such that $a_1$ =2 and $a_n - a_{n-1} = 2n$ for $n \geq 2$. Then $a_1 + a_2 + .. + a_{20}$ is equal to?

$a_1$ = 2
$a_2$ = 2 + 2x2
$a_3$ = 6 + 2x3
$a_4$ = 12 + 2x4
$a_5$ = 20 + 2x5

$a_n$ = $b_n$ + $g_n$ ,here $g_n$=2xn

$b_3$-$b_2$=4
$b_4$-$b_3$=6
$b_5$-$b_4$=8
$b_6$-$b_5$=10

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    $\begingroup$ Do list out a few terms to see the pattern. It is not too difficult to see the pattern. $\endgroup$ Commented Apr 16, 2013 at 10:22

3 Answers 3

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As $a_n-a_{n-1}=2n, a_n$ can be at most quadratic.

Let $a_n=An^2+Bn+C$ where $A,B,C$ are arbitrary constants

So, $2n=a_n-a_{n-1}=A(2n-1)+B=2An+B-A$

Comparing the coefficients of $n,2A=2, A=1$

Comparing the constants $B-A=0\implies B=A=1$

So, $a_n= n^2+n +C$

$2=a_1=2+C\implies C=0\implies a_n= n^2+n $

So, $$\sum_{1\le r\le n}a_r= \sum_{1\le r\le n}r^2+ \sum_{1\le r\le n}r= \frac{n(n+1)(2n+1)}6+ \frac{n(n+1)}2=\frac{n(n+1)(n+2)}3$$

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I have 2 methods of doing this:

Method one:

$a_2-a_1=4$, i.e. $a_2=6$

$a_3-a_2=6$, i.e. $a_3=12$

$a_4-a_3=8$, i.e. $a_4=20$

You will notice that $a_k = 2+4+6+8+ \cdots +2k = \frac{k}{2}(2k+2)=k^2 + k$ (using summation formula)

$a_1+a_2+ \cdots + a_{20} = \displaystyle\sum_{k=1}^{20} (k^2+k)=\frac{(20)(20+1)(2\times20+1)}{6}+20\times\frac{1+20}{2}=3080$

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Since $a_n-a_{n-1}=2n$, So we have $\displaystyle\sum^{i=20}_{i=2}a_n-a_{n-1}=\displaystyle\sum^{i=20}_{i=2}2n$.

Now we can calculate series. There fore, $a_{20}=418+2$ ,as $a_{20}-a_1=\displaystyle\sum^{i=20}_{i=2}2n$. By repeating this procedure you can find $a_1+a_2\cdots+a_n$.

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  • $\begingroup$ $a_{20}=418+2$. In any case, the OP was to find the sum $a_1+\cdots+a_{20}$. $\endgroup$ Commented Apr 16, 2013 at 12:28

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