Pulling a long molecule

The beads are touching each other, they are not connected by string. So, they have attraction forces, balanced by normal reaction.The force on first bead is clearly $F_{0} \dfrac{\pi^{2}}{12} \approx 8.225N$ So, if we apply this force, bead 1 will get free, but it might also be possible that before the breaking of bead 1, the first $n$ beads are broken from the rest.

Force on the system of first 2 beads due to others is

$F_{2} = F_{1} + F_{1} - F_{0} \approx 6.45 N$

Similarly, force on first 3 beads due to others is :

$F_{3} = F_{1} + F_{1} - F_{0} + F_{1} - F_{0} \bigg( 1 - 1/2^2\bigg) \approx 7.175 N$

We can easily deduce that force on first $n$ beads due to others is:

$F_{n} = \displaystyle n F_{1} - F_{0} \bigg( \sum_{r=1}^{n-1} \sum_{k=1}^{r} \dfrac{(-1)^{k-1} }{r^2}\bigg)$

= $\displaystyle nF_{0} - \sum_{r=1}^{n-1} \dfrac{(-1)^{r-1} (n-r)}{r^2}$

We get $F_{4} = 6.79, F_{5} = 7.022, F_{6} = 6.86$

$F_{n} = nF_{0} \displaystyle \sum_{r=1}^{\infty} \dfrac{(-1)^{r-1}}{r^2} - \sum_{r=1}^{n-1} \dfrac{(-1)^{r-1} (n-r)}{r^2}$

= $n \displaystyle \sum_{r=n}^{\infty} \dfrac{(-1)^{r-1}}{r^2} + F_{0} \displaystyle \sum_{r=1}^{n-1} \dfrac{(-1)^{r-1}}{r}$

Clearly, it converges to $F_{0} \ln 2 \approx 6.932 N$

The minimum force required is $F_{2}= 6.45 N$

I look on the beaded molecule as follows ....+-+-+-+-+-+ (arbitrarily assume + is at the end) The force acting between the last charged-bead on the right to the rest of the Molec. is (assuming an infinite long Molecule) 10x(pi^2)/12 which is about 8.2246N. There is also a repulsive force if we look at the molecule as follows .......+-+-+-+-+X(-)+ where we "deleted" one negative charge next to the end(marked X(-)). The repulsive force is per using the given serial formula, but omiting the said charge from the sum= 10((pi^2/12)-1)=1.7753N. Deducting the attracting force from the repulsive force will result at a bond strength of 6.4493N between X(-) and the successive + charge(#3 from the first). So X is pulled from one side with all the rest of the Molecule at 8.2246N, but including a repulsive force on the bond 3(+)X(-) will net 6.4493N. See it as follows 1.7753<X>8.2246=6.4493> still net attraction Same computation on other bonds will give less repulsive force because of the distance from the end of the molecule, assuming also that 8.2246 remains the same attraction along the Molec. because of infinite length. So 3(+)X(-) bond seems the weak link of the Molecule with a strength of 6.4493N.