Mathematics - Binomial Theorem Question with Solution | TestHub

By substituting $x=1$ and $x=-1$ into the equation and adding the results, we obtain:

$$a_0+a_2+\dots +a_{50}=\frac{3^{25}+1}{2}$$

We can expand $(1+2{)}^{25}$ using the binomial theorem:

$$(1+2{)}^{25}={}^{25}{\mathrm{C}}_0+{}^{25}{\mathrm{C}}_1\cdot 2+{}^{25}{\mathrm{C}}_2\cdot 2^2+\dots +{}^{25}{\mathrm{C}}_{25}\cdot 2^{25}$$

Therefore,

$$\frac{(1+2{)}^{25}+1}{2}=\frac{{}^{25}{\mathrm{C}}_0+{}^{25}{\mathrm{C}}_1\cdot 2+{}^{25}{\mathrm{C}}_2\cdot 2^2+\dots +{}^{25}{\mathrm{C}}_{25}\cdot 2^{25}+1}{2}$$

Since ${}^{25}{\mathrm{C}}_0=1$, the expression becomes:

$$\frac{1+{}^{25}{\mathrm{C}}_1\cdot 2+{}^{25}{\mathrm{C}}_2\cdot 2^2+\dots +{}^{25}{\mathrm{C}}_{25}\cdot 2^{25}+1}{2}$$

$$=\frac{2+{}^{25}{\mathrm{C}}_1\cdot 2+{}^{25}{\mathrm{C}}_2\cdot 2^2+\dots +{}^{25}{\mathrm{C}}_{25}\cdot 2^{25}}{2}$$

$$=1+{}^{25}{\mathrm{C}}_1+{}^{25}{\mathrm{C}}_2\cdot 2+\dots +{}^{25}{\mathrm{C}}_{25}\cdot 2^{24}$$

This result is an even integer because all terms after the first '1' are multiples of 2.