Total no of terms in the expansion of $(x+z)^{50} +(x-z)^{50}$

Question

Please help me in solving the question and also tell me the technique so that I can easily solve it when similar question come in the form of Multiple choice questions MCQS.

Answer 1

Individually, we expand the expressions.

$$(x+z)^{50} = \sum\limits_{k=0}^{50}\binom{50}{k}x^{50-k}z^k$$

$$(x-z)^{50}=(x+(-z))^{50}=\sum\limits_{k=0}^{50}\binom{50}{k}x^{50-k}(-z)^k$$

Each expansion comes from the binomial theorem.

Now, we notice that when $k$ is even, the coefficient of $x^{50-k}z^k$ from the first is equal to the coefficient of $x^{50-k}z^k$ of the second (since a negative number raised to an even power is positive). Further we notice that when $k$ is odd, the coefficients are of the same magnitude but are of opposite signs, and so when added will cancel out.

We have then

$$(x+z)^{50}+(x-z)^{50}=2\sum\limits_{k=0}^{25}\binom{50}{2k}x^{50-2k}z^{2k}$$

and in particular, this expression will have 26 (nonzero) terms.

Worded another way, the even indexed terms survive while the odd indexed terms disappear. The $26$ even numbered terms between $0$ and $50$ (inclusive) survive, while the $24$ odd numbered terms between $0$ and $50$ disappear.

Answer 2

Both $(x-z)^{50}$ and $(x+z)^{50}$ have $51$ terms after expanding them; however, those in which $z$ has an odd degree cancel out (do you understand why?), so we are left with $26$ terms.