Find $a,b,c$ such that $(2\cdot 1-1)+(2\cdot 2-1)+(2\cdot 3-1)+...+(2\cdot n-1)=an^2+bn+c$

Question

Can I get some hints?

Based on the pattern on the LHS, I think you meant the first term to be $(21-1)$ instead of $(21-n)$, unless you meant to put it like that. Also, "such as" should be "such that" in the question. — Aiden Chow, Aug 29 '20 at 19:11

score 1 · Answer 1 · answered Aug 29 '20 at 19:11

1

Hint 1: Try to substitute small values for $n$ (say $0,1$ and $2$) and make a system of 3 equations. Result prove with induction.

Hint 2: $$1+2+3+...+n = {n(n+1)\over 2}$$

answered Aug 29 '20 at 19:11

nonuser

90,026

Aiden Chow · Accepted Answer · 2020-08-29T19:49:41.027

1

The LHS is the sum of the first $n$ odd numbers, which results in $n^2$. So the equation becomes: $$n^2=an^2+bn+c$$in which $\boxed{a=1,b=c=0}$ follows(comparing coefficients).

edited Aug 29 '20 at 19:49

answered Aug 29 '20 at 19:16

Aiden Chow

2,849
10
32

score 0 · Answer 3 · answered Aug 29 '20 at 19:24

0

for $n=1$, $a+b+c=1$ for $n=2$, $4a+2b+c=4$ for $n=3$, $9a+3b+c=9$ $a=1, b=c=0$

answered Aug 29 '20 at 19:24

Lion Heart

7,073

Find $a,b,c$ such that $(2\cdot 1-1)+(2\cdot 2-1)+(2\cdot 3-1)+...+(2\cdot n-1)=an^2+bn+c$

3 Answers3