Properties of Newton's Binom Formula

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Properties for `(a+b)^n` are following:

  1. Number of all summands in expansion equals `n+1`.
  2. General member of expansion has form `T_(k+1)=C_n^ka^(n-k)b^k`, where k=0,1,2,...,n.
  3. Coefficients that are equidistant from ends of expansion are equal.
  4. Sum of all binomial coefficient is `2^n`.
  5. Sum of binomial coefficients on even positions equals sum of coefficients on odd positions.

Example 1. Find the greatest element in the expansion `(a+b)^n`, if sum of all binomial coefficients is 4096.

According to property 4, we have that `2^n=4096` or n=12.

Since 12 is even number than the greatest coefficient is near middle (i.e. `12/2=6`-th) member. This coefficient equals `C_(12)^6=(12!)/(6!(12-6)!)=(12!)/(6!6!)=(12*11*10*9*8*7*6!)/(6!6!)=924`.

Example 2. In expansion of `(sqrt(z)-2/(root(3)z))^15` find member that doens't contain z.

We can rewrite above expression as `(z^(1/2)+(-2z^(-1/3)))^15`. Here `a=z^(1/2),b=-2z^(-1/3),n=15`.

Now, using property 2, we have that `T_(k+1)=C_(15)^k(z^(1/2))^(15-k)(-2z^(-1/3))^k=C_(15)^k(-2)^kz^((15-k)/2-k/3)`.

Member will not contain z, when power of z equals 0, i.e. `(15-k)/2-k/3=0` or `k=9`.

Therefore, 10-th member doesn't contain z and it equals `T_(9+1)=C_(15)^9(-2)^9=-2562560`.

Example 3. Find the greatest binomial coefficient of expansion `(n+1/n)^n`, if product of forth member from the beginning and fourth member from the end equals 14400.

Fourth member from the beginning is `T_4=T_(3+1)=C_n^3n^(n-3)*1/n^3`, and fourth member from the end is `T_((n+1-4)-1)=T_(n-2)=C_n^(n-3)n^3 1/(n^(n-3))`.

Therefore, their product equals `T_4T_(n-2)=C_n^3C_n^(n-3)=C_n^3C_n^3=(C_n^3)^2=14400`. From this we have that `C_n^3=120`.

We can think a bit differently: from property 3 we have that coefficient are equal, coefficient near fourth member from beginning is `C_n^3`, therefore coefficient near fourth member from the end is also `C_n^3`; thus, their product is `C_n^3*C_n^3=14400` or `C_n^3=120`.

Now, `(n!)/((n-3)!3!)=120` or `(n(n-1)(n-2)(n-3)!)/((n-3)!6)=120`. This gives `n(n-1)(n-2)=720=10*9*8`, so n=10.

Therefore, the greatest coefficient is near 5-th member (`10/2=5`) and it equals `C_(10)^5=(10!)/(5!5!)=252`.