Can't solve numeric sequence problem!!!

downmath

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I'm currently doing some numeric sequence problems on an Algebra worksheet. But I've stumbled upon this problem and I don't seem to be able to solve it.

The problem is as follows:

The general term of (an)\left ( a_{n} \right ) is an=1(n+1)2a_{n} = \frac{1}{\left ( n+1 \right )^{2}} .

Find the first five terms of (bn)\left ( b_{n} \right ) , the general term of which is

bn=(an)2an1b_{n} = \left ( a_{n} \right )^{2}-a_{n-1} .

So far I've managed to find the first 6 terms of (an)\left ( a_{n} \right ) :


a1=14;a2=19;a3=116;a4=125;a5=136;a6=149;a_{1} = \frac{1}{4} ; a_{2} = \frac{1}{9} ; a_{3} = \frac{1}{16} ; a_{4} = \frac{1}{25} ; a_{5} = \frac{1}{36} ; a_{6} = \frac{1}{49} ; .

But I don't know how to go about finding (b1)\left ( b_{1} \right ) because I would need an1a_{n-1} which is (a0)\left ( a_{0} \right ). I've thought about finding (a0)\left ( a_{0} \right ) and then substituting it in (b1)\left ( b_{1} \right ) but I don't know if that is the right approach.
 
I'm currently doing some numeric sequence problems on an Algebra worksheet. But I've stumbled upon this problem and I don't seem to be able to solve it.

The problem is as follows:

The general term of (an)\left ( a_{n} \right ) is an=1(n+1)2a_{n} = \frac{1}{\left ( n+1 \right )^{2}} .

Find the first five terms of (bn)\left ( b_{n} \right ) , the general term of which is

bn=(an)2an1b_{n} = \left ( a_{n} \right )^{2}-a_{n-1} .

So far I've managed to find the first 6 terms of (an)\left ( a_{n} \right ) :


a1=14;a2=19;a3=116;a4=125;a5=136;a6=149;a_{1} = \frac{1}{4} ; a_{2} = \frac{1}{9} ; a_{3} = \frac{1}{16} ; a_{4} = \frac{1}{25} ; a_{5} = \frac{1}{36} ; a_{6} = \frac{1}{49} ; .

But I don't know how to go about finding (b1)\left ( b_{1} \right ) because I would need an1a_{n-1} which is (a0)\left ( a_{0} \right ). I've thought about finding (a0)\left ( a_{0} \right ) and then substituting it in (b1)\left ( b_{1} \right ) but I don't know if that is the right approach.
Given the situation:

THAT is the correct approach.
 
a0=1(0+1)2=1\displaystyle a_0= \frac{1}{(0+1)^2}= 1.

Nothing difficult about that!
 
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