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Worked solution

Separate an arithmetic sigma sum into two simple sums

Evaluate ∑k=1n(7k+5)\sum_{k=1}^{n}(7k+5) for positive integer nn.

CJC · Sequences and Series · Example 23

Answer

12n(7n+17)\frac12n(7n+17).

Compare your reasoning

Separate the linear and constant terms

∑k=1n(7k+5)=7∑k=1nk+∑k=1n5=7(1+2+⋯+n)+(5+5+⋯+5).\begin{aligned}\sum_{k=1}^{n}(7k+5)&=7\sum_{k=1}^{n}k+\sum_{k=1}^{n}5\\&=7(1+2+\cdots+n)+(5+5+\cdots+5).\end{aligned}

The natural-number sum is an AP with a=1,d=1 and n terms; the constant 5 is repeated n times.

Apply the two sum formulas and simplify

7[12n(n+1)]+5n=12n(7n+17)\begin{aligned}&7\left[\frac12n(n+1)\right]+5n\\ & =\frac12n(7n+17)\end{aligned}

This is the full printed simplification.

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