Binomial Theorem Question - None

Question

In the expansion of $(1-2\sqrt{x})^{50} = \displaystyle\sum_{r=0}^{50} {}^{50}C_r (1)^{50-r}(-2\sqrt{x})^r = \displaystyle\sum_{r=0}^{50} {}^{50}C_r(-2)^r (x)^{r/2}$, the sum of rational terms is:

Accepted Answer

Step 1: Identify when terms are rational. The general term is ${}^{50}C_r(-2)^r x^{r/2}$. For rationality, $r/2$ must be an integer, so r must be even.

Step 2: Let r = 2k where k = 0, 1, 2, ..., 25. Rational terms are ${}^{50}C_{2k}(-2)^{2k}x^k = {}^{50}C_{2k}(4)^k x^k$.

Step 3: Sum of rational terms = $\sum_{k=0}^{25} {}^{50}C_{2k} \cdot 4^k \cdot x^k$. Setting x = 1 gives the sum of coefficients: $\sum_{k=0}^{25} {}^{50}C_{2k} \cdot 4^k$.

Step 4: Use the identity: $(1+y)^{50} + (1-y)^{50} = 2\sum_{k=0}^{25} {}^{50}C_{2k}y^{2k}$. Setting y = 2 (so $y^{2k} = 4^k$):

$(1+2)^{50} + (1-2)^{50} = 2\sum_{k=0}^{25} {}^{50}C_{2k} \cdot 4^k$

$3^{50} + (-1)^{50} = 2\sum_{k=0}^{25} {}^{50}C_{2k} \cdot 4^k$

$3^{50} + 1 = 2\sum_{k=0}^{25} {}^{50}C_{2k} \cdot 4^k$

∴ Sum of rational terms = $\frac{3^{50}+1}{2}$

Question:

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