4

There is a list of numbers $a_{1} , a_{2} , …, a_{2010}$ . For $1 \leq n \leq 2010$, where $n$ is positive integer, let $a_1+a_2+ \ldots +a_n = S_n$ . If $a_1 = 2010$ and $S_n = a_nn^2$ for all n, what is the value of $a_{2010}$ ?

I've been trying to manipulate the formula but I cant seem to find a good relationship between $a_1$ and $a_{2010}$ like $$ a_{2010} = \frac{a_1 +a_2 ... +a_{2010}}{2010^2} $$ Then tried to use the definition $S_n = a_nn^2 $ over and over again but I cant find a good formula.

Elias Costa
  • 14,658
SuperMage1
  • 2,486

6 Answers6

2

We have

$$S_2=a_2n^2=4a_2=a_1+a_2 \implies a_2=\frac13 a_1=670$$

$$S_3=a_3n^2=9a_3=a_1+a_2+a_3 \implies 8a_3=\frac43 a_1 \implies a_3=\frac16 a_1=335$$

$$S_4=a_4n^2=16a_4=a_1+a_2+a_3+a_4 \implies 15a_4=\frac32 a_1 \implies a_3=\frac1{10} a_1=201$$

then we claim that $$a_n=\frac1{T(n)}a_1=\frac{2}{n(n+1)}a_1$$ to be proved by induction, that is

$$S_{n+1}=a_{n+1}(n+1)^2=S_n+a_{n+1}=a_nn^2+a_{n+1}\implies a_{n+1}((n+1)^2-1)=\frac{2n^2}{n(n+1)}a_1$$

$$\implies a_{n+1}=\frac{2n^2}{n(n+1)(n^2+2n)}a_1=\frac{2}{(n+1)(n+2)}a_1$$

user
  • 154,566
2

Start from the fact that (for $n>1$) $a_n=S_n-S_{n-1}$. This means $$a_n=a_nn^2-a_{n-1}(n-1)^2.$$ If you use this to express $a_n$ in terms of $a_{n-1}$, and substitute in the corresponding expression for $a_{n-1}$, etc, a pattern should emerge.

Especially Lime
  • 40,828
2

Since $a_n=\frac{1}{n^2-1}\sum_{k=1}^{n-1}a_k$, the strong induction hypothesis $a_n=\frac{2a_1}{n(n+1)}$ gives $$a_n=\frac{2a_1}{(n-1)(n+1)}\sum_{k=1}^{n-1}(\frac{1}{k}-\frac{1}{k+1})=\frac{2a_1}{(n-1)(n+1)}(1-\frac{1}{n})=\frac{2a_1}{n(n+1)}.$$With our inductive proof complete, $a_{2010}=\frac{2}{2011}$.

J.G.
  • 115,835
2

$$S_n = a_n n^2$$ $$a_{n}+S_{n-1}=a_nn^2$$

$$S_{n-1}=a_n(n^2-1)$$

$$a_{n-1}(n-1)^2=a_n(n^2-1)$$

$$a_{n-1}(n-1)=a_n(n+1)$$

$$a_n=\frac{n-1}{n+1}a_{n-1}=\frac{n-1}{n+1}\frac{n-2}{n}a_{n-2}=\frac{n-1}{n+1}\frac{n-2}{n}\frac{n-3}{n-1}a_{n-3}=\frac{(2)(1)}{(n+1)n}a_1$$

Siong Thye Goh
  • 149,520
  • 20
  • 88
  • 149
2

Hint:

$$S_n-S_{n-1}=a_n$$ and $$n^2a_n-(n-1)^2a_{n-1}=a_n$$

so that

$$a_n=\frac{n-1}{n+1}a_{n-1}.$$

Then

$$a_n=\frac{n-1}{n+1}a_{n-1}=\frac{n-1}{n+1}\frac{n-2}{n-0}a_{n-2}=\frac{n-1}{n+1}\frac{n-2}{n-0}\frac{n-3}{n-1}a_{n-3}=\frac{n-1}{n+1}\frac{n-2}{n-0}\frac{n-3}{n-1}\frac{n-4}{n-2}a_{n-4}=\cdots$$

More generally, after simplification,

$$a_n=\frac2{n+1}\frac1{n-0}a_1=2\left(\frac1n-\frac1{n+1}\right)a_1$$ and the sum telescopes.

1

We have $a_n\cdot n^2=a_1+a_2+\ldots+a_n=S_n$. Then $$ \begin{array}{cc} \begin{array}{rlrl} \frac{\quad}{\quad}a_1\cdot 1^2=&2010\\ \frac{\quad}{\quad}a_2\cdot 2^2=&a_1+a_2\\ \frac{\quad}{\quad}a_3\cdot 3^2=&a_1+a_2+a_3\\ \frac{\quad}{\quad}a_4\cdot 4^2=&a_1+a_2+a_3+a_4\\ \end{array} & \begin{array}{rl} \implies a_1=&2010\\ \implies a_2 =& \frac{1}{2^2-1}2010\\ \implies a_3=&\frac{1}{3^2-1}2010+\frac{1}{3^2-1}\frac{1}{2^2-1}2010\\ \implies a_4=&\frac{1}{4^2-1}2010+\frac{1}{4^2-1}\frac{1}{3^2-1}2010+\frac{1}{4^2-1}\frac{1}{3^2-1}\frac{1}{2^2-1}2010\\ \end{array} \end{array} $$ Now is easy to see $$ a_n=2010\left(\frac{1}{n^2-1}+\frac{1}{n^2-1}\cdot\frac{1}{(n-1)^2-1}+\ldots+\frac{1}{n^2-1}\cdot\frac{1}{(n-1)^2-1}\cdot\ldots\cdot\frac{1}{2^2-1}\right) $$

Elias Costa
  • 14,658