What is the f(n)=(11^{n+1}-4^{n+1})\mod 7 ?

The f(n)=(11^{n+1}-4^{n+1})\mod 7 is

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f(n)=(11^{n+1}-4^{n+1})\mod 7

f (n) = (1 1^{n + 1} - 4^{n + 1}) mod 7

Domain : - \infty < n < \infty

X Intercepts : (- 1, 0), Y Intercepts : (0, 49 mod)

Asymptotes : None

Extreme Points : None

Interval Notation

Domain : (- \infty, \infty)

Solution steps

Domain of

(1 1^{n + 1} - 4^{n + 1}) mod 7 : - \infty < n < \infty

Axis interception points of

(1 1^{n + 1} - 4^{n + 1}) mod 7 :

X Intercepts

: (- 1, 0),

Y Intercepts

: (0, 49 mod)

Asymptotes of

(1 1^{n + 1} - 4^{n + 1}) mod 7 :

None

Extreme Points of

(1 1^{n + 1} - 4^{n + 1}) mod 7 :

None

\frac{2 x ^{3} + x ^{2} + 8 x + 6}{( x ^{2} + 4 ) ( x ^{2} + 1 )}

\frac{6 ( k + 126 ) ( - k ^{2} + 1001 k - 132930 )}{k - 4238}

x (2 lo g_{10} (x) + 1)

- 4 (m + 6 = - 8)

\frac{2 x ^{2} - 4 x + 9}{( 2 - x ) ( x + 3 )}

What is the f(n)=(11^{n+1}-4^{n+1})\mod 7 ?
The f(n)=(11^{n+1}-4^{n+1})\mod 7 is