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

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

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

expand

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

na + \frac{n ^{2} d}{2} - \frac{n d}{2} + \frac{n}{2} + \frac{1}{2}

Solution steps

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

Remove parentheses:

(a) = a

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

Expand

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

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

Expand

n (2 a + n d - d) + n + 1 : 2 na + n^{2} d - n d + n + 1

= \frac{2 na + n ^{2} d - n d + n + 1}{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 + n + 1}{2} = \frac{2 na}{2} + \frac{n ^{2} d}{2} - \frac{n d}{2} + \frac{n}{2} + \frac{1}{2}

= \frac{2 na}{2} + \frac{n ^{2} d}{2} - \frac{n d}{2} + \frac{n}{2} + \frac{1}{2}

Divide the numbers:

\frac{2}{2} = 1

= na + \frac{n ^{2} d}{2} - \frac{n d}{2} + \frac{n}{2} + \frac{1}{2}

y (x - ln (y))

(1, 69 - 1, 21 i) 0.36 i

(y sin (x) + sin (y) + y) d x

\frac{20 + 3 x}{2 ( 15 + x )}

\frac{b}{x ^{2} \cdot ( x + a )}

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