If the sum of odd terms is A and the sum of e | vedclass

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(C) The binomial expansion is given by: $(x + a)^n = {^nC_0}x^n + {^nC_1}x^{n-1}a + {^nC_2}x^{n-2}a^2 + {^nC_3}x^{n-3}a^3 + \dots$Let $A$ be the sum o

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$4AB = (x + a)^{2n} - (x - a)^{2n}$

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(C) The binomial expansion is given by: $(x + a)^n = {^nC_0}x^n + {^nC_1}x^{n-1}a + {^nC_2}x^{n-2}a^2 + {^nC_3}x^{n-3}a^3 + \dots$Let $A$ be the sum of odd terms (1st,3rd,5th,...): $A = {^nC_0}x^n + {^nC_2}x^{n-2}a^2 + {^nC_4}x^{n-4}a^4 + \dots$Let $B$ be the sum of even terms (2nd,4th,6th,...): $B = {^nC_1}x^{n-1}a + {^nC_3}x^{n-3}a^3 + {^nC_5}x^{n-5}a^5 + \dots$We know that $(x + a)^n = A + B$ and $(x - a)^n = A - B$.Multiplying these two expressions: $(x + a)^n(x - a)^n = (A + B)(A - B) = A^2 - B^2$.However,the question asks for the relationship involving $4AB$. Note that $(A + B)^2 - (A - B)^2 = 4AB$.Substituting the binomial expressions: $(x + a)^{2n} - (x - a)^{2n} = 4AB$.

If the sum of odd terms is $A$ and the sum of even terms is $B$ in the expansion of $(x + a)^n$,then:

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