If (1 + x + x^2)^{25} = a_0 + a_1x + a_2x^2 + | vedclass

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(A) Let $f(x) = (1 + x + x^2)^{25} = a_0 + a_1x + a_2x^2 + ..... + a_{50}x^{50}$.Putting $x = 1$,we get $f(1) = (1 + 1 + 1)^{25} = 3^{25} = a_0 + a_1

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even

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(A) Let $f(x) = (1 + x + x^2)^{25} = a_0 + a_1x + a_2x^2 + ..... + a_{50}x^{50}$.Putting $x = 1$,we get $f(1) = (1 + 1 + 1)^{25} = 3^{25} = a_0 + a_1 + a_2 + ..... + a_{50}$.Putting $x = -1$,we get $f(-1) = (1 - 1 + 1)^{25} = 1^{25} = 1 = a_0 - a_1 + a_2 - ..... + a_{50}$.Adding the two equations,we get $f(1) + f(-1) = 3^{25} + 1 = 2(a_0 + a_2 + a_4 + ..... + a_{50})$.Therefore,$a_0 + a_2 + a_4 + ..... + a_{50} = \frac{3^{25} + 1}{2}$.Since $3^{25}$ is odd,$3^{25} + 1$ is even,so the sum is an integer.We can write $3^{25} + 1 = (1 + 2)^{25} + 1 = (1 + ^{25}C_1(2) + ^{25}C_2(2^2) + ..... + 2^{25}) + 1$.$= 2 + 2(^{25}C_1 + ^{25}C_2(2) + ..... + 2^{24}) = 2[1 + ^{25}C_1 + ^{25}C_2(2) + ..... + 2^{24}]$.Dividing by $2$,we get $1 + ^{25}C_1 + ^{25}C_2(2) + ..... + 2^{24}$,which is clearly an even number.

If $(1 + x + x^2)^{25} = a_0 + a_1x + a_2x^2 + ..... + a_{50}x^{50}$,then $a_0 + a_2 + a_4 + ..... + a_{50}$ is :

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