Answer:
The statement is true for every n between 0 and 77 and it is false for [tex]n\geq 78[/tex]
Step-by-step explanation:
First, observe that, for n=0 and n=1 the statement is true:
For n=0: [tex]\sum^{n}_{i=0} (2i)^4=0 \leq 0=(4n)^4[/tex]
For n=1: [tex]\sum^{n}_{i=0} (2i)^4=16 \leq 256=(4n)^4[/tex]
From this point we will assume that [tex]n\geq 2[/tex]
As we can see, [tex]\sum^{n}_{i=0} (2i)^4=\sum^{n}_{i=0} 16i^4=16\sum^{n}_{i=0} i^4[/tex] and [tex](4n)^4=256n^4[/tex]. Then,
[tex]\sum^{n}_{i=0} (2i)^4 \leq(4n)^4 \iff \sum^{n}_{i=0} i^4 \leq 16n^4[/tex]
Now, we will use the formula for the sum of the first 4th powers:
[tex]\sum^{n}_{i=0} i^4=\frac{n^5}{5} +\frac{n^4}{2} +\frac{n^3}{3}-\frac{n}{30}=\frac{6n^5+15n^4+10n^3-n}{30}[/tex]
Therefore:
[tex]\sum^{n}_{i=0} i^4 \leq 16n^4 \iff \frac{6n^5+15n^4+10n^3-n}{30} \leq 16n^4 \\\\ \iff 6n^5+10n^3-n \leq 465n^4 \iff 465n^4-6n^5-10n^3+n\geq 0[/tex]
and, because [tex]n \geq 0[/tex],
[tex]465n^4-6n^5-10n^3+n\geq 0 \iff n(465n^3-6n^4-10n^2+1)\geq 0 \\\iff 465n^3-6n^4-10n^2+1\geq 0 \iff 465n^3-6n^4-10n^2\geq -1\\\iff n^2(465n-6n^2-10)\geq -1[/tex]
Observe that, because [tex]n \geq 2[/tex] and is an integer,
[tex]n^2(465n-6n^2-10)\geq -1 \iff 465n-6n^2-10 \geq 0 \iff n(465-6n) \geq 10\\\iff 465-6n \geq 0 \iff n \leq \frac{465}{6}=\frac{155}{2}=77.5[/tex]
In concusion, the statement is true if and only if n is a non negative integer such that [tex]n\leq 77[/tex]
So, 78 is the smallest value of n that does not satisfy the inequality.
Note: If you compute [tex](4n)^4- \sum^{n}_{i=0} (2i)^4[/tex] for 77 and 78 you will obtain:
[tex](4n)^4- \sum^{n}_{i=0} (2i)^4=53810064[/tex]
[tex](4n)^4- \sum^{n}_{i=0} (2i)^4=-61754992[/tex]
