Sum of n n avanced Part 2

Algebra Level pending

We know that Gauss calculate ∑ k = 1 n k = n ( n + 1 ) 2 \displaystyle \sum_{k=1}^n k=\frac{n(n+1)}{2}

But what if we write in form of a + ( a + b ) + ( a + 2 b ) + ( a + 3 b ) + ( a + 4 b ) + ( a + 5 b ) + ⋯ + ( a + n b ) a+(a+b)+(a+2b)+(a+3b)+(a+4b)+(a+5b)+\dots+(a+nb) then what does it equal to?

( n + 1 ) 2 ( 2 a + n b ) \frac{(n+1)}{2(2a+nb)} ( 2 a − n b ) ( n + 1 ) 2 \frac{(2a-nb)(n+1)}{2} ( 2 a + n b ) ( n + 1 ) 2 \frac{(2a+nb)(n+1)}{2}

Gia Hoàng Phạm by Gia Hoàng Phạm , 19, Vietnam

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3 solutions

S = a + ( a + b ) + ( a + 2 b ) + ⋯ + ( a + n b ) = a + ∑ k = 1 n ( a + k b ) = a + a ∑ k = 1 n 1 + b ∑ k = 1 n k = a + n a + b n ( n + 1 ) 2 = a ( n + 1 ) + b n ( n + 1 ) 2 = ( 2 a + n b ) ( n + 1 ) 2 \begin{aligned} S & = a + (a+b) + (a+2b) + \cdots + (a+nb) \\ & = a + \sum_{k=1}^n (a+kb) \\ & = a + a \sum_{k=1}^n 1 + b \sum_{k=1}^n k \\ & = a + na + \frac {bn(n+1)}2 \\ & = a(n+1) + \frac {bn(n+1)}2 \\ & = \boxed{\dfrac {(2a+nb)(n+1)}2} \end{aligned}

3 Helpful 0 Interesting 0 Brilliant 0 Confused
Brian Moehring
Oct 3, 2018

Add the sum with itself in reversed order: S = a + ( a + b ) + ⋯ + ( a + ( n − 1 ) b ) + ( a + n b ) ( n + 1 terms ) + S = ( a + n b ) + ( a + ( n − 1 ) b ) + ⋯ + ( a + b ) + a ( n + 1 terms ) 2 S = ( 2 a + n b ) + ( 2 a + n b ) + ⋯ + ( 2 a + n b ) + ( 2 a + n b ) ( n + 1 terms ) \begin{array}{cccccccccccccc} & S & = & a & + & (a+b) & + & \cdots & + & (a+(n-1)b) & + & (a+nb) & \qquad & \left(n+1 \text{ terms}\right) \\ + & S & = & (a+nb) & + & (a+(n-1)b) & + & \cdots & + & (a+b) & + & a & & \left(n+1 \text{ terms}\right) \\ \hline \\ & 2S & = & (2a+nb) & + & (2a+nb) & + & \cdots & + & (2a+nb) & + & (2a+nb) & & \left(n+1 \text{ terms}\right) \end{array}

Therefore, since 2 S 2S is the sum of n + 1 n+1 copies of 2 a + n b , 2a+nb, we have 2 S = ( 2 a + n b ) ( n + 1 ) , 2S = (2a+nb)(n+1), and S = ( 2 a + n b ) ( n + 1 ) 2 S = \frac{(2a+nb)(n+1)}{2}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

a + ( a + b ) + ( a + 2 b ) + ( a + 3 b ) + ( a + 4 b ) + ( a + 5 b ) + ⋯ + ( a + n b ) = a ( n + 1 ) + n b ( n + 1 ) 2 = ( 2 a + n b ) ( n + 1 ) 2 a+(a+b)+(a+2b)+(a+3b)+(a+4b)+(a+5b)+\dots+(a+nb)=a(n+1)+\frac{nb(n+1)}{2}=\frac{(2a+nb)(n+1)}{2}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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