The question asks to simplify the following expression by writing it as a single summation (the second sigma in the expression should have upper limit of n+2):

$$3\sum_{i=2}^n (2i^2-i) - \sum_{j=4}^n (3j^2+1)$$

Im having trouble figuring this one out, I tried to write each inner term as a sum then use recursive definitions to expand them all then simplify:

$$ \sum_{i=2}^n (6i^2) - \sum_{i=2}^n (3i) - \sum_{j=4}^n (3j^2) + \sum_{j=4}^n 1$$

$$6(\frac{n(n+1)(2n+1)}{6}) - 3(\frac{n(n+1)}{2}) - 3(\frac{(n+2)((n+2)+1)(2(n+1)+1)}{6} + \frac{n(n+1)}{2}$$

from here I continued to simplify until it was in the simplest form but i dont understand how to express it a single summation, also I think I may have gone about this the wrong way and made some mistakes here and there. Is anyone able to correct me or point me in the right direction in terms of solving this?


  • $\begingroup$ In your bold comment, are you saying that the upper limit should be $n+2$ in the beginning, or that it should be $n+2$ after the simplification? $\endgroup$
    – B. Goddard
    Commented Sep 23, 2017 at 12:12
  • $\begingroup$ $$n+2$$ in the beginning $\endgroup$
    – kr1s
    Commented Sep 24, 2017 at 6:46

3 Answers 3


(Assuming the upper limit is $n+2.$) You can change variables in a summation just like in an integral. Here, you want the limits to match, but the second sum's limits are $2$ larger than the first. So in the second sum set $i=j+2$ so that $j=i-2$. Then when $j=4$, $i=2$ and when $j=n+2$, $i=n$. The second sum becomes

$$\sum_{i=2}^{n} (3(i+2)^2 +1).$$

Now you can combine the sums and simplify the summand.

If the upper limit is not $n+2$, then you start by doing the same thing, but then there's some more work.

  • $\begingroup$ Thanks! this clears up so much misconception $\endgroup$
    – kr1s
    Commented Sep 24, 2017 at 7:22

Hint: Note that in the first derived expression there is a minus instead of a plus. \begin{align*} - \sum_{j=4}^n (3j^2+1)=- \sum_{j=4}^n 3j^2\color{blue}{-} \sum_{j=4}^n 1 \end{align*} It is often more convenient to simplify the sum at first and then applying summation formulas to the terms in the sum which are left (see (3) below).

Here is one of many different ways to simplify the sums (boldface variant).

\begin{align*} \color{blue}{3\sum_{j=2}^n}&\color{blue}{(2i^2-i)-\sum_{j=4}^{n+2}(3j^2+1)}\\ &=3\sum_{i=2}^n(2i^2-i)-\sum_{i=2}^{n}(3(i+2)^2+1)\tag{1}\\ &=\sum_{i=2}^n(6i^2-3i)-\sum_{i=2}^n(3i^2+12i+13)\tag{2}\\ &=\sum_{i=2}^n(3i^2-15i-13)\tag{3}\\ &=\frac{1}{2}n(n+1)(2n+1)-\frac{15}{2}n(n+1)-13n-(3-15-13)\tag{4}\\ &=\left(n^3+\frac{3}{2}n^2+\frac{1}{2}n\right)-\left(\frac{15}{2}n^2+\frac{15}{2}\right)-13n+26\tag{5}\\ &=\color{blue}{n^3-6n^2-20n+25}\tag{6} \end{align*}


  • In (1) we denote the index in the right-hand sum with $i$ and shift the index to start with $i=2$.

  • In (2) we mulitply out.

  • In (3) we collect the terms in one sum and are now well prepared to use the summation formulas.

  • In (4) we apply the summation formulas and subtract $(3-15-13)$ to compensate for the missing index value $i=1$.

  • In (5) we multiply out.

  • In (6) we do some final simplifications.

  • $\begingroup$ thankyou for this explanation! so can line (6) be written in sigma notation or does single summation just imply writing it as a simplified expression? $\endgroup$
    – kr1s
    Commented Sep 24, 2017 at 7:28
  • $\begingroup$ @kris: Here it is more convenient to write the summation as it is done in (6). Otherwise the term sigma-notation should explicitly occur. $\endgroup$ Commented Sep 24, 2017 at 7:40
  • $\begingroup$ ah ok, also how come the 13 in line (4) gets multiplied by n? $\endgroup$
    – kr1s
    Commented Sep 24, 2017 at 7:51
  • $\begingroup$ @kris: In (3) the term $13$ is part of the sum from $i=2$ to $i=n$. We obtain $(n-1)$ times $13$. The last summand in (4) compensates for $1\cdot 13$. $\endgroup$ Commented Sep 24, 2017 at 7:56

This may be an unnecessary answer if you actually want to have every simplification step: but I can give you a formula for the sum.

We start off by changing the second summation with the $n+2$ upper limit, B.Goddard explained how, so we get:

$$3\sum_{i=2}^{n} (2i^2 - i) - \sum_{i=2}^{n} (3(i+2)^2 +1)$$

Now the sum of a general quadratic sequence is the following (I have a proof of this on my computer somewhere but I won't bother going through that here): $$S_2(n,s,d_1,c) = \frac{n(cn^2 + 2c + 3nd_1 + 6s - 3cn - 3d_1)}{6}$$

Where S$_2(n,s,d_1,c)$ is the quadratic sum function, $n$ is the number of terms to calculate the sum for, $s$ the starting term of the quadratic sequence, $d_1$ is the first difference between the second term and the first term in the sequence and $c$ is the constant difference between the differences.

Putting in values for your sigmas we get that the sum equals to:

$$ (18-49) + (45-76) + (84 - 109) + \ ... \ = -31 + -31 + -25 + \ ... $$

So we see that $s= -31$, the first difference $d_1 = -31 - (-31) = 0$, and we get $c$ by seeing what the second difference is, $-25$, then subtracting the first difference from that, giving $c = -25 -(-31) = 6$. Remember that we calculate $n-1$ terms here, not $n$, because both summations starts at $2$. Input these values into the function and we get:

$$S_2(n-1,-31,0,6) = (n-1)(n^2 - 5n - 25)$$

Check it for yourself and see that it works!

  • $\begingroup$ i have no doubt that your answer is correct although i think quadratic sum function is a bit beyond what i have learnt so far in level 1 discrete math course. Thanks anyway, i will go and look it up in text book and see if i can make sense of it $\endgroup$
    – kr1s
    Commented Sep 24, 2017 at 7:53

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