What is expand (n(n+1)(2n+1))/6+3((n(n+1))/2) ?

The solution to expand (n(n+1)(2n+1))/6+3((n(n+1))/2) is (n^3)/3+2n^2+(5n)/3

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expand (n(n+1)(2n+1))/6+3((n(n+1))/2)

expand

\frac{n ( n + 1 ) ( 2 n + 1 )}{6} + 3 (\frac{n ( n + 1 )}{2})

\frac{n ^{3}}{3} + 2 n^{2} + \frac{5 n}{3}

Solution steps

\frac{n ( n + 1 ) ( 2 n + 1 )}{6} + 3 (\frac{n ( n + 1 )}{2})

Remove parentheses:

(a) = a

= \frac{n ( n + 1 ) ( 2 n + 1 )}{6} + 3 \cdot \frac{n ( n + 1 )}{2}

\frac{n ( n + 1 ) ( 2 n + 1 )}{6} = \frac{2 n ^{3} + 3 n ^{2} + n}{6}

3 \cdot \frac{n ( n + 1 )}{2} = \frac{3 n ^{2} + 3 n}{2}

= \frac{2 n ^{3} + 3 n ^{2} + n}{6} + \frac{3 n ^{2} + 3 n}{2}

Least Common Multiplier of

6, 2 : 6

Adjust Fractions based on the LCM

= \frac{2 n ^{3} + 3 n ^{2} + n}{6} + \frac{( 3 n ^{2} + 3 n ) \cdot 3}{6}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{2 n ^{3} + 3 n ^{2} + n + ( 3 n ^{2} + 3 n ) \cdot 3}{6}

Factor

2 n^{3} + 3 n^{2} + n + (3 n^{2} + 3 n) 3 : n \cdot 2 (n^{2} + 6 n + 5)

= \frac{n \cdot 2 ( n ^{2} + 6 n + 5 )}{6}

Cancel the common factor:

2

= \frac{n ( n ^{2} + 6 n + 5 )}{3}

Expand

n (n^{2} + 6 n + 5) : n^{3} + 6 n^{2} + 5 n

= \frac{n ^{3} + 6 n ^{2} + 5 n}{3}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{n ^{3} + 6 n ^{2} + 5 n}{3} = \frac{n ^{3}}{3} + \frac{6 n ^{2}}{3} + \frac{5 n}{3}

= \frac{n ^{3}}{3} + \frac{6 n ^{2}}{3} + \frac{5 n}{3}

Divide the numbers:

\frac{6}{3} = 2

= \frac{n ^{3}}{3} + 2 n^{2} + \frac{5 n}{3}

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What is expand (n(n+1)(2n+1))/6+3((n(n+1))/2) ?
The solution to expand (n(n+1)(2n+1))/6+3((n(n+1))/2) is (n^3)/3+2n^2+(5n)/3