$\left(\sum\limits_{i=1}^N (a_i+b_i)^2\right)^2 \stackrel{?}{=} \sum\limits_{i=1}^N (a_i+b_i)^4 $ where $a_i,b_i \in \mathbb{R}$

Question

Is $\left(\sum\limits_{i=1}^N (a_i+b_i)^2\right)^2 \text{equal to} \sum\limits_{i=1}^N (a_i+b_i)^4 $ where $a_i,b_i \in \mathbb{R}$ ??

If not then how to expand: $\left(\sum\limits_{i=1}^N (a_i+b_i)^2\right)^2$

Thanks

Actually this is true if and only if $a_i=-b_i$ for all $i$. — Lost1, Dec 30 '13 at 22:51
Weird question. I suspect OP didn't ask the real question he intended to ask. — Adam Rubinson, Mar 28 '23 at 14:44

score 5 · Answer 1 · answered Dec 30 '13 at 22:37

5

No. Take $a_i=2, b_i=0$. Then $16N^2\ne 16N$.

answered Dec 30 '13 at 22:37

Boris Novikov

score 2 · Answer 2 · answered Dec 30 '13 at 22:38

2

Hint: Let $u_i=(a_i+b_i)^2$, and notice that $(\sum_{i}u_i)^2=\sum_{i,j}u_iu_j$ then you go from there.

answered Dec 30 '13 at 22:38

bankrip

If $(\sum_{i}u_i)^2=\sum_{i,j}u_iu_j$, that means $i$ may not be equal to $j$. In that case it will not be equal to $\left(\sum\limits_{i=1}^N (a_i+b_i)^2\right)^2$. – kaka Dec 30 '13 at 23:01
1

what do you mean? i and j are indices of the summation. Double sum actually. – bankrip Dec 30 '13 at 23:08

score 1 · Answer 3 · answered Mar 28 '23 at 14:40

1

\begin{equation*} [\sum_{i=1}^N(a_i+b_i)^2]^2 = \sum_{i=1}^N (a_i+b_i)^4 + 2 \sum_{i=1}^N \sum_{j=1}^{i-1} (a_i+b_i)^2(a_j+b_j)^2 \end{equation*}

answered Mar 28 '23 at 14:40

温泽海

3 Answers3