There’s a list

5n+14-11n-2

-4(n+8)

10(n-5)

-11-10n+2n+1

Solution 1

Inspecting a 7th-degree polynomial of 'n' by taking its derivatives

From the given conditions,

 2n7+n6+2n4+2n3+2n2

=(n+1)7+(n+1)5+(n+1)4+2(n+1)3+(n+1)2+2(n+1)+1.

 2n7+n6+2n4+2n3+2n2

=n7+7n6+21n5+35n4+35n3+21n2+7n+1

 +n5+5n4+10n3+10n2+5n+1

 +n4+4n3+6n2+4n+1

 +2(n3+3n2+3n+1)

 +(n2+2n+1)+2(n+1)+1.

n7-6n6-22n5-39n4-49n3-42n2-26n-9=0 [1].

Let A(n)=n7-6n6-22n5-39n4-49n3-42n2-26n-9.

For n>2, A(n)<n7-6n6-22n5.

Since A(n)=0,

0<n7-6n6-22n5.

0<n5(n-3+√31)(n-3-√31).

Since n>2, n>3+√31, which means n≥9 [2].

Now, to differentiate A(n),

A'(n)=7n6-36n5-110n4-156n3-147n2-84n-26.

A''(n)=42n5-180n4-440n3-468n2-294n-84.

A'''(n)=210n4-720n3-1320n2-936n-294.

A(4)(n)=840n3-2160n2-2640n-936.

 A(5)(n)

=2520n2-4320n-2640

=2520{(n-6/7)2-262/147}.

If A(5)(n)=0, n=6/7±√(262/147).

Therefore, A(5)(n)>0 for n≥3.

Although n≥9 according to [2], I try inspecting A(n) for n≥10.

When n≥10, A(4)(n) increases monotonously.

A(4)(10)=840000-216000-26400-936>0.

Therefore, A(4)(n)>0 for n≥10.

And when n≥10, A'''(n) increases monotonously.

A'''(10)=2100000-720000-132000-9360-294>0.

Therefore, A'''(n)>0 for n≥10.

And when n≥10, A''(n) increases monotonously.

A''(10)=4200000-1800000-440000-46800-2940-84>0.

Therefore, A''(n)>0 for n≥10.

And when n≥10, A'(n) increases monotonously.

 A'(10)

=7000000-3600000-1100000-156000-14700-840-26>0.

Therefore, A'(n)>0 for n≥10.

And when n≥10, A(n) increases monotonously.

 A(10)

=107-6000000-2200000-390000-49000-4200-260-9>0.

Therefore, A(n)>0 for n≥10 [3].

From [2] and [3], A(n)=0 only when 9≤n<10.

In consequence, n=9.

the volume is the product of the area of the base, 3x(3x+4), and the height, 3x-1

step-by-step explanation:

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step-by-step explanation:

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