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enter image description here$(1+x+x^2)^{1061}=a_0+a_1x+...+a_{2122}x^{2122}$ then what is the value of $(1-a_1^{2}+a_2^{2}-a_3^{2}...)$ in terms of single $a_n$ ? n lies between 0 and 2122 . how to get in terms on $a_n$ ? Hints and suggestions please!Help!

See question number 24.

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    I don't get the question. The expression $(1-a_1^2+a_2^2\dots)$ (does it continue alternating?) already is in terms of the $a_i$. Can you clarify what you're asking for? – Christoph Apr 02 '15 at 09:08
  • yes it is an alternating expression and the value of the expression is equal to a particular coefficient of x's power in the expansion $(1+x+x^2)^{1061}$ –  Apr 02 '15 at 09:36
  • Could pascal's triangle or binomial theorem help here? –  Apr 02 '15 at 09:46
  • Where did you get in touch with the problem? – AD - Stop Putin - Apr 02 '15 at 09:59
  • Yeah..but that would'nt be the right technique to to use here as it is too long! –  Apr 02 '15 at 09:59
  • I came across this in my school's previous year's question paper...will i scan it and upload? @AD. –  Apr 02 '15 at 10:00

1 Answers1

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So you have $$(1+x+x^2)^{1061} = a_0 + a_1 x + a_2x^2+\cdots+a_{2122}x^{2122} \tag{1}$$ $$\implies x^{2122}(1-\tfrac1x+\tfrac1{x^2})^{1061} = a_0x^{2122} - a_1 x^{2121} + a_2x^{2120}-\cdots+a_{2122} \tag{2}$$

Multiplying $(1)$ and $(2)$, from the product of RHSs we get that $a_0^2-a_1^2+a_2^2\cdots+a_{2122}^2$ is the coefficient of $x^{2122}$ in the expansion of $LHS_1\cdot LHS_2=(1+x^2+x^4)^{1061}$, which is the same as the coefficient of $x^{1061}$ in the expansion of $(1+x+x^2)^{1061}$, which is $a_{1061}$.

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