Indexing with sigma notation

Problem Statement

Given

$$\sum_{k=1}^{n}(3k-1)=119,$$

find $n$.

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Hints

Split the sum into $3\sum k - \sum 1$.

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Solutions

$$\sum_{k=1}^{n}(3k-1)=3\cdot\frac{n(n+1)}{2}-n=\frac{3n^2+n}{2}.$$

Set equal to $119$:

$$\frac{3n^2+n}{2}=119 \implies 3n^2+n-238=0.$$ $$\Delta=1+2856=2857.$$

This is not a perfect square, so there is no integer $n$ with exact value $119$.

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Further Readings

• HSC Polys Ext 1: https://vumaths.com/booklets/hsc-polys-ext-1/
• HSC Combinatorics: https://vumaths.com/booklets/hsc-combinatorics/
• HSC Vectors: https://vumaths.com/booklets/hsc-vectors/
• HSC Trigonometry: https://vumaths.com/booklets/hsc-trigonometry/

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