Find the biggest term of the expansion of $\left( \frac{2}{3} + \frac{1}{3} \right)^{100}$.
Here is what I've tried:
Every term of the expansion has the following form: $$T_{k+1} = \binom{100}{k} \left( \frac{2}{3} \right)^{100 - k} \left( \frac{1}{3} \right)^k$$
This form can be rewritten in two ways: $$T_{k+1} = \binom{100}{k} \frac{1}{2^k} \left( \frac{2}{3} \right)^{100}$$ $$T_{k+1} = \binom{100}{k} 2^{100-k} \left( \frac{1}{3} \right)^{100}$$
This is where I got stuck.
Thank you in advance!
For the largest term in the expansion T(k) such that [T(k+1)/T(k)]>1...now just plug in the values of the general terms in this formula to get the value of k.
In your case we get the value of k as 33...which means that the largest term in the given expansion is the 34th term.
A shortcut formula to find out the largest term no. in a given binomial expansion of the form (a+b)^n is as follows:
k <= (n+1)b/(a+b)
The required term no. will then be equal to (k+1).
In case the variables are with negative sign....you can use the above formula by taking only their absolute values(mod)....Also take greatest integer function of k when you get decimal values.....hope this helps!!