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JEE Main & Advanced JEE Main Paper (Held On 9 April 2014)

  • question_answer
    The number of terms in the expansion of\[{{(1+x)}^{101}}{{(1+{{x}^{2}}-x)}^{100}}\]in powers of x is:   [JEE Main Online Paper ( Held On 09 Apirl  2014  )

    A) 302                                        

    B) 301

    C) 202                                        

    D) 101

    Correct Answer: C

    Solution :

                    Given expansion is \[{{(1+x)}^{101}}{{(1-x+{{x}^{2}})}^{100}}\] \[=(1+x){{(1-x)}^{100}}{{(1-x+{{x}^{2}})}^{100}}\] \[=(1+x){{[(1-x)(1-x+{{x}^{2}})]}^{100}}\] \[=(1+x)[{{(1-{{x}^{3}})}^{100}}]\] Expansion \[={{(1+{{x}^{3}})}^{100}}\]will have 100 + 1 = 101 terms So, \[=(1+x){{(1-{{x}^{3}})}^{100}}\]will have \[2\times 101=202\] terms


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