I used equation $$(AL_1 )^2 + (L_1M_1)^2 +(AL_12 )^2 + (L_2M_2)^2 + (AL_3 )^2 + (L_3M_3)^2 + ......=(a^2 + 1^2 +a^2 + 2^2 +a^2 + 3^2 +a^2 + 4^2 + .......) + ((a-1)^2+1^2 +(a-2)^2+2^2 +(a-3)^2+3^2 +....)$$

So its general term is $$(a)^2 + r^2 + (a-r)^2 + r^2$$

Applying sigma $$\sum_{r=1}^{a-1} (a)^2 + r^2 + (a-r)^2 + r^2$$ I resolved this into $$2(a^2) + 3(r^2) -2ar$$ and applied sigma$$\sum_{r=1}^{a-1} 2(a^2) + 3(r^2) -2ar$$ I am having problem in applying sigma please anybody help me please explain in detail since i am still learning about sigma application

• Aug 13, 2018 at 16:41
• Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. Aug 13, 2018 at 16:44
• @mzp sir please edit -1 in applying sigma line it is actually (a-1) as upper limit of sigma and also later equations are wrong Aug 13, 2018 at 16:56
• Just put $a-1$ in curly brackets, i.e. $\{a-1\}$. (I am not allowed to edit just that.)
$\sum_\limits{r=1}^{a-1} (2(a^2) + 3(r^2) -2ar)\\ \sum_\limits{r=1}^{a-1} 2a^2 + \sum_\limits{r=1}^{a-1}3r^2 -\sum_\limits{r=1}^{a-1}2ar\\ 2a^2(a-1) + 3\frac {(a)(a-1)(2a-1)}{6}-2a\frac {a(a-1)}{2}$
• I would expect a solution to be cubic, as we have the square of the length of 2(a-1) lines. So, I am doubtful that the answer could be $\frac 12 (a-1)(4a-1)$ Aug 14, 2018 at 15:15