Solve the sum $\sum _{i=1}^4\:\sum _{j=1}^i\:\left(i\cdot \:j-1\right)$

Question

Is it correct if this sum is solved this way?

Answer 1

Yes, it looks good. You can also check your work on wolfram alpha. Go to wolframalpha.com and enter:

sum i = 1 to 4 sum j = 1 to i (i*j - 1)

as in here.

Answer 2

It looks as if you have used the formulas for the sum of a square and a cube, or something. That's great.

But I can never remember those (so I wouldn't be able to tell you if you did it correctly or not). I only remember the Gauß formula, $\sum_{i=1}^ni=\frac{n(n+1)}2$ (which you did use at one point).

However, since the sum is only up to $4$, you could just do it by hand.

Thus, picking up in the middle, $\sum_{i=1}^4\frac12(i^3+i^2)-i=\frac12(1^3+1^2+2^3+2^2+3^3+3^2+4^3+4^2)-\frac{4\cdot 5}2=\frac12(1+1+8+4+27+9+64+16)-10=\frac12(130)-10=65-10=55$.

So, either you lucked out, or you used the formulas correctly.

Answer 3

$$\begin{align} \sum_{i=1}^n\sum_{j=1}^i (ij-1) &=\sum_{i=1}^ni\left(\sum_{j=1}^ij-1\right)\\ &=\sum_{i=1}^ni\left(\frac {i(i+1)}2-1\right)\\ &=\frac 12\sum_{i=1}^ni(i^2+i-2)\\ &=\frac 12 \sum_{i=1}^n i(i-1)(i+2)\\ &=\frac 12 \sum_{i=1}^n(i+1)i(i-1)+i(i-1)\\ &=\frac 12 \sum_{i=1}^n 6\binom {i+1}3+2\binom i2\\ &=3\binom {n+2}4+\binom {n+1}3\\ &=\binom{n+1}3\left(3\cdot \frac {n+2}4+1\right)\\ &=\frac 14 \binom {n+1}3(3n+10)\\ &=\frac 1{24}(n-1)n(n+1)(3n+10)\end{align}$$ Alternatively,

$$\begin{align} \sum_{i=1}^n\sum_{j=1}^i (ij-1) &=\frac 12\left(\sum_{i=1}^ni^3+\sum_{i=1}^ni^2\right)-\sum_{i=1}^n i\\ &=\frac 12 \left[\binom {n+1}2 ^2+\frac 13\binom {n+1}2 (2n+1)-2\binom {n+1}2\right]\\ &=\frac 12\binom {n+1}2 \left[\binom {n+1}2+\frac 13 (2n+1)-2\right]\\ &=\frac 12 \binom {n+1}2\cdot \frac16 \big[3n(n+1)+2(2n+1)-12\big]\\ &=\frac 1{12} \binom {n+1}2\cdot (3n^2+7n-10)\\ &=\frac 1{12}\binom {n+1}2 (n-1)(3n+10)\\ &=\frac 14\binom{n+2}3(3n+10)\\ &=\frac 1{24}(n-1)n(n+1)(3n+10) \end{align}$$