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Home » In the expansion of if the sum of odd terms is denoted by and the sum of even term by . Then prove that(i) (ii)
In the expansion of (x+a)^{n} if the sum of odd terms is denoted by O and the sum of even term by \mathrm{E}. Then prove that
(i) O^{2}-E^{2}=\left(x^{2}-a^{2}\right)^{n}
(ii) 4 O E=(x+a)^{2 n}-(x-a)^{2 n}
Binomial Theorem | CBSE | Class 11 | Maths | NCERT Exemplar
In the expansion of (x+a)^{n} if the sum of odd terms is denoted by O and the sum of even term by \mathrm{E}. Then prove that
(i) O^{2}-E^{2}=\left(x^{2}-a^{2}\right)^{n}
(ii) 4 O E=(x+a)^{2 n}-(x-a)^{2 n}

(i) We know,

(x+a)^{n}=

{ }^{n}C_{0} x^{n}+{ }^{n} C_{1} x^{n-1} a^{1}+{ }^{n} C_{2} x^{n-2} a^{2}+{ }^{n} C_{3} x^{n-3} a^{3}+\ldots

Sum of odd terms will be,

O={ }^{n} C_{0} x^{n}+{ }^{n} C_{2} x^{n-2} a^{2}+\ldots

And the sum of even terms will be,

E={ }^{n} C_{1} x^{n-1} a+{ }^{n} C_{3} x^{n-3} a^{3}+\ldots

Since (x+a)^{n}=0+E so,

(x-a)^{n}=0-E

As a result,

(\mathrm{O}+\mathrm{E})(\mathrm{O}-\mathrm{E})=(\mathrm{x}+\mathrm{a})^{\mathrm{n}}(\mathrm{x}-\mathrm{a})^{\mathrm{n}}

(ii) 40 \mathrm{E}=(\mathrm{O}+\mathrm{E})^{2}-(\mathrm{O}-\mathrm{E})^{2}

=(x-a)^{2 n}-(x-a)^{2 n}

i

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