मराठी

In the expansion of (x + a)n if the sum of odd terms is denoted by O and the sum of even term by E. Then prove that O2 – E2 = (x2 – a2)n

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प्रश्न

In the expansion of (x + a)n if the sum of odd terms is denoted by O and the sum of even term by E. Then prove that O2 – E2 = (x2 – a2)n 

बेरीज
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उत्तर

Given expression is (x + a)n

(x + a)n = nC0xn a0 + nC1xn–1a + nC2xn–2a2 + nC3xn–3a3 + … + nCnan

Sum of odd terms,

O = `""^n"C"_0 x^n + ""^n"C"_2 x^(n - 2)a^2 + ""^n"C"+4x^(n - 4)a^4` + ...

And the sum of even terms,

E = `""^n"C"_1x^(n - 1) * a + ""^n"C"_3x^(n - 3)a^3 + ""^n"C"_5x^(n - 5)a^5` + ...

Now (x + a)n = O + E  ......(i)

Similarly (x – a)n = O – E   .....(ii)

Multiplying equation (i) and equation (ii), we get,

(x + a)n (x – a)n = (O + E)(O – E)

⇒ (x2 – a2)n = O2 – E2

Hence O2 – E2 = (x2 – a2)n

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पाठ 8: Binomial Theorem - Exercise [पृष्ठ १४३]

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एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 11
पाठ 8 Binomial Theorem
Exercise | Q 15.(i) | पृष्ठ १४३

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