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Question
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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Solution
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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