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Now, this may be a very easy problem but I came across this in an examination and I could not solve it.

Find the value of

$$3+7+12+18+25+\ldots=$$

Now here is my try

$$3+7+12+18+25+\ldots=\\3+(3+4)+(3+4+5)+(3+4+5+6)+(3+4+5+6+7)+\ldots=\\3n+4(n-1)+5(n-2)+\ldots$$

After that, I could not proceed.

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  • $\begingroup$ What do you mean by "..."? $\endgroup$ – Chappers Apr 12 '17 at 15:30
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    $\begingroup$ I think the question is find a formula for the finite sum, otherwise 3+7+12+18+25+... = infinity. $\endgroup$ – Dimitris Apr 12 '17 at 15:32
  • $\begingroup$ yeah correct @dimitris $\endgroup$ – Pole_Star Apr 12 '17 at 15:32
  • $\begingroup$ Ok, we have two different answers.... $\endgroup$ – Pole_Star Apr 12 '17 at 15:40
  • $\begingroup$ Gregory's answer is correct (even though he has much less rep than Ross) :-) $\endgroup$ – bubba Apr 12 '17 at 15:43
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Although I prefer Gregory's answer that computes this directly, here is another approach:

Notice:

$s_1 = 3$; $s_2 = 10$; $s_3 = 22$; $s_4 = 40$; $s_5 = 65$

Let $s_n = an^3 + bn^2 + cn + d$

Now, solve the system of equations given by

$s_1 = 3$; $s_2 = 10$; $s_3 = 22$; $s_4 = 40$;

to find that:

$a = 1/6, b = 3/2, c = 4/3, d = 0$, hence:

$s_n = 1/6n^3 + 3/2n^2 + 4/3n$

which yields the same as Gregory's answer.

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    $\begingroup$ what justifies supposing it's cubic? $\endgroup$ – Dimitris Apr 12 '17 at 15:49
  • $\begingroup$ This is possibly more direct, but doesn't this require the person to know that the sum will be cubic? Though I suppose if you knew the terms were quadratic, you could make the leap that their sum has to be cubic... $\endgroup$ – Gregory Apr 12 '17 at 15:49
  • $\begingroup$ Trial and error. I first supposed it was quadratic, then cubic, ..... Usually, when such questions are given, the degree will not be too high. $\endgroup$ – user370967 Apr 12 '17 at 15:50
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    $\begingroup$ Your approach is good...sort of unorthodox $\endgroup$ – Pole_Star Apr 12 '17 at 16:00
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    $\begingroup$ @dp1611 you can find the common-difference to determine whether it's a linear, quadratic, cubic, etc, or exponential. Using that technique, your sequence will have a common-difference of 1 after doing it 3 times (aka. cubic) $\endgroup$ – Andrew T. Apr 12 '17 at 16:44
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The above sequence is given by $a_i = \frac{i(i+5)}{2}$. The finite sum is given by \begin{align} S_n & = \sum_{i=1}^n a_i \\ & = \frac{1}{2} \sum_{i=1}^n i^2 + \frac{5}{2} \sum_{i=1}^n i \\ & = \frac{n(n+1)(2n+1)}{12} + \frac{5n(n+1)}{4} \\ & = \frac{n(n+1)(n+8)}{6} \end{align}

EDIT: (In response to comments) To a certain point, finding the formula for $a_i$ is trial and error. However, it is not difficult to note here that the difference of the difference of terms is always 1 (i.e. $\Delta^2 a_i = 1$). This implies that the dependence is quadratic. Using $a_i = A i^2 + B i + C$ we can determine the formula with three terms.

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  • $\begingroup$ How did you find the expression for the sequence in the first place? $\endgroup$ – jjmontes Apr 12 '17 at 17:00
  • $\begingroup$ my question also but hit and trial may be the answer. $\endgroup$ – Pole_Star Apr 12 '17 at 19:45
  • $\begingroup$ much clear after the edit $\endgroup$ – Pole_Star Apr 14 '17 at 6:26
  • $\begingroup$ One can also solve the recurrence relation $a_{n+1} - a_n = n+3$ to find $a_n$ $\endgroup$ – user370967 Apr 14 '17 at 14:21
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    $\begingroup$ yes, more or less the same idea I think. $\endgroup$ – Gregory Apr 14 '17 at 20:03
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I'm adding another answer because people ask how to find $a_n$ without trial and error.

We note that:

$a_2 - a_1 = 4; a_3 - a_2 = 5; a_4 - a_3 = 6$,

which leads us to conclude that $a_n$ is given by the recurrence relation:

$a_{n+1} - a_n = n+3$

Let's start by solving the homogeneous equation:

$a_{n+1} - a_n = 0$

The associated polynomial is $P(r) =r - 1$. The root of this polynomial is $r = 1$. Therefore, a solution to the homogeneous equation is $a_n^{h} = 1^n = 1$.

Now, we want to find a particular solution, so let's try $a_n^{p} = An + B$.

Then:

$A(n+1) + B - An - B= n+3$. This does not work.

Let's try $a_n^{p} = An^2 + Bn + C$

Thus:

$A(n+1)^2 + B(n+1) + C - An^2 - Bn - C = n +3$

from which follows:

$2A = 1, A + B = 3, C \in \mathbb{R}$.

Hence:

$A = 1/2; B = 5/2; C \in \mathbb{R}$

Therefore, $a_n^p = 1/2n^2 + 5/2n + C$

and

$a_n = a_h + a_p = 1 + 1/2n^2 + 5/2n + C = 1/2n^2 + 5/2n + D$

Because $a_1 = 3$, it follows that $D = 0$.

We conclude that $a_n = 1/2n^2 + 5/2n \quad \triangle$

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If you know the value of $n$ then $$3+7+12+18+25+\ldots=\\3n-3n+3+7+12+18+25+\ldots$$ As $3=1+2$ we can write as $$\\1+2+3+(1+2+3+4)+(1+2+3+4+5)+(1+2+3+4+5+6)+\ldots-3n=\\\left(\sum_{n=3}^n{\frac{n(n+1)}{2}}\right)-3n$$

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  • $\begingroup$ I think your answer is wrong. It does not match with Gregory's answer. $\endgroup$ – Pole_Star Apr 13 '17 at 12:11
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This answer is unique in that it does not use trial and error to find out the expression for the $n$-th term.

Our sum is given by :

$$ S=3+7+12+18+...+T_n \\ S=\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;3+7+12+18+...+T_n-1+T_n $$ Subtracting these two sums : $$ 0=3+4+...+(n+2)-T_n \\ T_n=3+4+...+(n+2) \\ T_n=\frac{(n+2)(n+3)}{2}-3 =\frac{n(n+5)}{2} $$

Now, you can use Gregory's answer to figure the sum out.

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