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Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem
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Find $(x+1)^6 + (x – 1)^6$. Hence, evaluate $( \sqrt2 + 1)^6+(\sqrt 2 –1)^6$.

$\begin{array}{1 1}198 \\ 150 \\ 200 \\ 308 \end{array} $

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1 Answer

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Toolbox:
  • Express the given number as the sum or difference of two numbers whose powers are easier to evaluate. Then use binomial theorem as follows:
  • $(a+b)^n = \large \sum \limits_{k=0}^{n}\; $$^n \large C$$_k \; a^{n-k}b^k =\; ^n \large C$$_0\;a^nb^0 +\; ^n \large C$$_1\;a^{n-1}b^1+.... ^n\large C$$_n\;a^0b^n$, where $b^0 = 1 = a^{n-n}$
We can now use Binomial Theorem to evaluate $(x+1)^6 + (x – 1)^6.$
$(x+1)^6 = \;^6\large C$$_0\;x^61^0 + \;^6\large C$$_1\;x^51^1+....$
$(x-1)^6 = \;^6\large C$$_0\;x^61^0 - \;^6\large C$$_1\;x^51^1+....$
$(x+1)^6 + (x – 1)^6 = \;^6\large C$$_0\;x^61^0 + \;^6\large C$$_1\;x^51^1+.... + ( \;^6\large C$$_0\;x^61^0 - \;^6\large C$$_1\;x^51^1+....)$
$\qquad = 2 ( \;^6\large C$$_0\;x^6 + \;^6\large C$$_2\;x^4 + \;^6\large C$$_4\;x^2 + \;^6\large C$$_6)$
$\qquad = 2 (x^6+15x^4+15x^2+1)$
Therefore $( \sqrt2 + 1)^6+(\sqrt 2 –1)^6 = 2 (\sqrt{2}^6+15\;\sqrt{2}^4+15\;\sqrt{2}^2+1)$
$\qquad = 2(8+60+30+1)$
$\qquad = 2 (99)$
$\qquad = 198$
answered Apr 10, 2014 by balaji.thirumalai
edited Apr 10, 2014 by balaji.thirumalai
 

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