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Let $\{a_{n}\},\{b_{n}\}$ be sequences such that $$\sqrt{\dfrac{a_{n}+1}{b_{n}}}=\dfrac{1}{n}, \quad S_{n}=\sqrt{T_{n-1}} \quad(n>1)$$ where $\displaystyle S_{n}=\sum_{i=1}^{n}a_{i}$, $\displaystyle T_{n}=\sum_{i=1}^{n}b_{i}$, and $b_{1}=1$.

Find $T_{20}-S_{20}$.

(a) $15980$

(b) $35910$

(c) $43910$

(d) $43920$

My try: Since $$n^2(a_{n}+1)=b_{n},S^2_{n}=T_{n-1}$$ so $$S^2_{n+1}-S^2_{n}=a_{n+1}[S_{n+1}+S_{n}]=T_{n}-T_{n-1}=b_{n}.$$ Then I don't know how to continue. Thank you.

I guess this $$1^3+2^3+\cdots+n^3=(1+2+\cdots+n)^2$$ maybe usefull,so I guess $$a_{n}=n-1,b_{n}=n^3$$ and after I find this is true,But I can't prove it( without Mathematical induction)

math110
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1 Answers1

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We'll prove that $a_n=(n-1)$ and $b_n=n^3$ by complete induction.

For $n=1$, we have $b_1=1$ straight from problem statement and $a_1=0$ follows from $\sqrt{\frac{a_1+1}{b_1}}=1$.

For any higher $n$, we have $$S_n=\sum_{k=1}^n a_k=\left(0+1+\ldots+(n-2)\right)+a_n=\frac{(n-1)(n-2)}{2}+a_n$$ and $$T_{n-1}=\sum_{k=1}^{n-1} b_k = \left(1^3 + 2^3 + \ldots + (n-1)^3\right) = \left(\frac{n(n-1)}{2}\right)^2$$

Equating $S_n^2 = T_{n-1}$ gives us $$a_n^2 + 2\frac{(n-1)(n-2)}{2}a_n + \left(\frac{(n-1)(n-2)}{2}\right)^2 = \left(\frac{n(n-1)}{2}\right)^2$$ which, after simplification, yields $$\left(a_n - (n-1)\right)\left(a_n + (n-1)^2\right) = 0$$

This gives us two possibilities: $a_n = (n-1)$ or $a_n = -(n-1)^2$. The latter would yield negative $S_n$, though, and that is impossible due to $S_n$ being equal to a square root; leaving the former as the only option.

$b_n$ can be obtained directly from $\sqrt{\frac{a_n+1}{b_n}}=\frac{1}{n}$; giving us $b_n=n^3$ as expected.

Now that we know the expressions for $a_n$ and $b_n$, it is straightforward to calculate $S_n=\frac{1}{2}n(n-1)$ and $T_n=\frac{1}{4}\left(n(n+1)\right)^2$ and $T_n-S_n=\frac{1}{4}n(n^3 + 2n^2 - n + 2)$ so $T_{20}-S_{20}=43910$.

Peter Košinár
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  • Thank you, I think this problem have other methods, – math110 Feb 17 '14 at 15:18
  • @chinamath By which method you want to get to the solution. – Singh Feb 22 '14 at 14:08
  • @Peter Košinár Sir, I think you are wrong in the starting and so your whole solution is wrong it is clearly given in problem that n>1 but you have started with n=1. You are much more experience than me so please give reason if I'm wrong. – Singh Feb 22 '14 at 14:23
  • @Singh The way I'm interpreting the problem statement is that the inequality $n>1$ applies (only) to the preceding equality $S_n=\sqrt{T_{n-1}}$ (since for $n=1$ it'd refer to $T_0$; something that actually turns out to be true, but is unnecessary) – Peter Košinár Feb 22 '14 at 18:30