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Algebra Difficulty 2.7 Junior Find the answer

If xk+1=xk+12x_{k+1} = x_k + \frac12 for k=1,2,,n1k=1, 2, \dots, n-1 and x1=1x_1=1, find x1+x2++xnx_1 + x_2 + \dots + x_n:

Pick one

Solution

The sequence x1,x2,,xnx_1, x_2, \dots, x_n is an arithmetic sequence since every term is 12\frac12 more than the previous term. Letting a=1a=1 and r=12r=\frac12, we can rewrite the sequence as a,a+r,,a+(n1)ra, a+r, \dots, a+(n-1)r.
Recall that the sum of the first nn terms of an arithmetic sequence is na+(n2)rna+\binom{n}2r.
Substituting our values for aa and rr, we get n+(n2)2n+\frac{\binom{n}2}{2}. Simplifying gives
n2+3n4 (E)\boxed{\frac{n^2+3n}{4}\textbf{ (E)}}

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.