What is (k(2k-1)(2k+1))/3+(2(k+1)-1)^2 ?

The solution to (k(2k-1)(2k+1))/3+(2(k+1)-1)^2 is (k(2k-1)(2k+1))/3+4k^2+4k+1

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simplify (k(2k-1)(2k+1))/3+(2(k+1)-1)^2

simplify

\frac{k ( 2 k - 1 ) ( 2 k + 1 )}{3} + (2 (k + 1) - 1)^{2}

\frac{k ( 2 k - 1 ) ( 2 k + 1 )}{3} + 4 k^{2} + 4 k + 1

Solution steps

\frac{k ( 2 k - 1 ) ( 2 k + 1 )}{3} + (2 (k + 1) - 1)^{2}

Expand

2 (k + 1) - 1 : 2 k + 1

= \frac{k ( 2 k - 1 ) ( 2 k + 1 )}{3} + (2 k + 1)^{2}

Expand

(2 k + 1)^{2} : 4 k^{2} + 4 k + 1

= \frac{k ( 2 k - 1 ) ( 2 k + 1 )}{3} + 4 k^{2} + 4 k + 1

\frac{9}{2} - 1

\frac{3}{b ^{2} - 4 b - 5} - \frac{2}{b ^{2} - 6 b + 5}

72 x + 15 x^{2} - 3 x^{3}

x^{2} - 3 x - 2 = 0

23 x + 5 > - 110

What is (k(2k-1)(2k+1))/3+(2(k+1)-1)^2 ?
The solution to (k(2k-1)(2k+1))/3+(2(k+1)-1)^2 is (k(2k-1)(2k+1))/3+4k^2+4k+1