Help plzz

ummm it still looks weird... you should use the LaTeX button on the toolbar on the top! :)

sorry, this problem looks super hard :( my small brain can't handle this haha 

Ok\( (a +(2n+1)d)^2- (a + (2n)d)^2 +(a + (2n-1)d)^2 - (a+(2n-2)d)^2 + \cdots + (a+d)^2 - a^2.\)

Can't you just use a calculator?

Answered here: https://web2.0calc.com/questions/reposted-because-the-answer-i-got-last-time-was-incorrect#r9

Compute the sum:
\((a +(2n+1)d)^2- (a + (2n)d)^2 +(a + (2n-1)d)^2 - (a+(2n-2)d)^2 + \cdots + (a+d)^2 - a^2\).

\(\small{ \begin{array}{|rcll|} \hline && \mathbf{\Big(a +(2n+1)d\Big)^2- \Big(a + (2n)d\Big)^2 +\Big(a + (2n-1)d\Big)^2 - \Big(a+(2n-2)d\Big)^2 + \cdots + (a+d)^2 - a^2} \\\\ &=& \Big(a +(2n+1)d\Big)^2 +\Big(a + (2n-1)d\Big)^2+\Big(a + (2n-3)d\Big)^2 + \cdots + \Big(a+(2n-(2n-1))d\Big)^2 \\ && -\Big(a + (2n-0)d\Big)^2- \Big(a+(2n-2)d\Big)^2- \Big(a+(2n-4)d\Big)^2 - \cdots -\Big(a+(2n-(2n))d\Big)^2 \\\\ &=& \mathbf{\sum \limits_{k=1}^{n+1}\Big(~ a+(2k-1)d~\Big)^2 -\sum \limits_{k=1}^{n+1}\Big(~ a+2(k-1)d~\Big)^2} \\ \hline \end{array} }\)

\(\begin{array}{|rcll|} \hline && \mathbf{\sum \limits_{k=1}^{n+1}\Big(~ a+(2k-1)d~\Big)^2 -\sum \limits_{k=1}^{n+1}\Big(~ a+2(k-1)d~\Big)^2} \\\\ &=& \sum \limits_{k=1}^{n+1}\Big(~ a+(2k-1)d~\Big)^2 - \Big(~a+2(k-1)d~\Big)^2 \\\\ &=& \sum \limits_{k=1}^{n+1} a^2+2ad(2k-1)+d^2(2k-1)^2-\Big(~ a^2+4ad(k-1)+4d^2(k-1)^2 ~\Big) \\\\ &=& \sum \limits_{k=1}^{n+1} {\color{red}{a^2}}+2ad(2k-1)+d^2(2k-1)^2 {\color{red}{-a^2}}-4ad(k-1)-4d^2(k-1)^2 \\\\ &=& \sum \limits_{k=1}^{n+1} 2ad(2k-1)+d^2(2k-1)^2 -4ad(k-1)-4d^2(k-1)^2 \\\\ &=& \sum \limits_{k=1}^{n+1} {\color{red}{4adk}}-2ad+d^2(2k-1)^2 {\color{red}{-4adk}}+4ad-4d^2(k-1)^2 \\\\ &=& \sum \limits_{k=1}^{n+1} -2ad+d^2(2k-1)^2 +4ad-4d^2(k-1)^2 \\\\ &=& \sum \limits_{k=1}^{n+1} 2ad+d^2(2k-1)^2 -4d^2(k-1)^2 \\\\ &=& \sum \limits_{k=1}^{n+1}2ad +d^2~\sum \limits_{k=1}^{n+1} (2k-1)^2 -4(k-1)^2 \quad | \quad \sum \limits_{k=1}^{n+1}2ad = 2ad(n+1) \\\\ &=& 2ad(n+1) +d^2\sum \limits_{k=1}^{n+1} (2k-1)^2 -4(k-1)^2 \\\\ &=& 2ad(n+1) +d^2\sum \limits_{k=1}^{n+1} 4k^2-4k+1-4(k^2-2k+1) \\\\ &=& 2ad(n+1) +d^2\sum \limits_{k=1}^{n+1} {\color{red}{4k^2}}-4k+1{\color{red}{-4k^2}}+8k-4 \\\\ &=& 2ad(n+1) +d^2\sum \limits_{k=1}^{n+1} 4k-3 \\\\ &=& 2ad(n+1) +d^2\Big(~4\sum \limits_{k=1}^{n+1}k-\sum \limits_{k=1}^{n+1}3~\Big)\quad | \quad \sum \limits_{k=1}^{n+1}3 = 3(n+1) \\\\ &=& 2ad(n+1) +d^2\Big(~4\sum \limits_{k=1}^{n+1}k-3(n+1)~\Big) \\\\ &=& 2ad(n+1) +d^2\Big(~4\sum \limits_{k=1}^{n+1}k-3(n+1)~\Big) \quad | \quad \sum \limits_{k=1}^{n+1}k = \dfrac{1+(n+1)}{2}(n+1) \\\\ &=& 2ad(n+1) +d^2\Big(~4\dfrac{(n+2))}{2}(n+1)-3(n+1)~\Big) \\\\ &=& 2ad(n+1) +d^2\Big(~2(n+2)(n+1)-3(n+1)~\Big) \\\\ &=& 2ad(n+1) +d^2(n+1)\Big(~2(n+2)-3~\Big) \\\\ &=& 2ad(n+1) +d^2(n+1)(2n+4-3) \\\\ &=& \mathbf{2ad(n+1) +d^2(n+1)(2n+1)} \\ \hline \end{array}\)

\(\mathbf{\Big(a +(2n+1)d\Big)^2- \Big(a + (2n)d\Big)^2 +\Big(a + (2n-1)d\Big)^2 - \Big(a+(2n-2)d\Big)^2 + \cdots + (a+d)^2 - a^2} \\= \mathbf{2ad(n+1) +d^2(n+1)(2n+1)}\)