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ArbitrLikvidat [17]
4 years ago
6

Show that sum k=1/2n(n+1) from k=1 to n

Mathematics
1 answer:
Setler79 [48]4 years ago
4 0
Suppose the value of the sum is S:

S=\displaystyle\sum_{k=1}^nk

So

S=1+2+3+\cdots+(n-2)+(n-1)+n

but also

S=n+(n-1)+(n-2)+\cdots+3+2+1

That is,

S=\displaystyle\sum_{k=1}^n(n-k+1)

Adding these together, we have

2S=\displaystyle\sum_{k=1}^n(k+n-k+1)=\sum_{k=1}^n(n+1)=(n+1)\sum_{k=1}^n1=n(n+1)
\implies S=\dfrac{n(n+1)}2
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