What is expand (n(2a+d(n-1)))/2 ?

The solution to expand (n(2a+d(n-1)))/2 is na+(n^2d)/2-(nd)/2

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expand (n(2a+d(n-1)))/2

expand

\frac{n ( 2 a + d ( n - 1 ) )}{2}

na + \frac{n ^{2} d}{2} - \frac{n d}{2}

Solution steps

\frac{n ( 2 a + d ( n - 1 ) )}{2}

Expand

2 a + d (n - 1) : 2 a + n d - d

= \frac{n ( n d - d + 2 a )}{2}

Expand

n (2 a + n d - d) : 2 na + n^{2} d - n d

= \frac{2 na + n ^{2} d - n d}{2}

Apply the fraction rule:

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

\frac{2 na + n ^{2} d - n d}{2} = \frac{2 na}{2} + \frac{n ^{2} d}{2} - \frac{n d}{2}

= \frac{2 na}{2} + \frac{n ^{2} d}{2} - \frac{n d}{2}

Divide the numbers:

\frac{2}{2} = 1

= na + \frac{n ^{2} d}{2} - \frac{n d}{2}

(x + x^{2} + 2 e^{2 x})^{2}

\frac{- y ( k - 2 ) - 11}{6}

4 (u y) y (u d y + y d u)

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

e^{(x^{2} + 1)^{3}}

What is expand (n(2a+d(n-1)))/2 ?
The solution to expand (n(2a+d(n-1)))/2 is na+(n^2d)/2-(nd)/2