Using Binomial Theorem, prove that 6n - 5n always leaves remain
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    Binomial Theorem

    Multiple Choice QuestionsShort Answer Type

    11.

    Find left parenthesis straight a plus straight b right parenthesis to the power of 4 minus left parenthesis straight a minus straight b right parenthesis to the power of 4. then evaluate space space left parenthesis square root of 3 plus square root of 2 right parenthesis to the power of 4 minus left parenthesis square root of 3 minus square root of 2 right parenthesis to the power of 4

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    12.

    Find left parenthesis straight x plus 1 right parenthesis to the power of 5 plus left parenthesis straight x minus 1 right parenthesis to the power of 5 and hence evaluate left parenthesis square root of 2 plus 1 right parenthesis to the power of 5 plus space left parenthesis square root of 2 minus 1 right parenthesis to the power of 5

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    13.

    Simplify: left parenthesis straight x plus 2 straight y right parenthesis to the power of 10 plus left parenthesis straight x minus 2 straight y right parenthesis to the power of 10

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    14. Show that 9n + 1 - 8n - 9 is divisible by 64, whenever n is a positive integer. 
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    15. Using Binomial Theorem, prove that 6n - 5n always leaves remainder 1 when divided by 25.

    6 to the power of straight n space equals space left parenthesis 1 plus 5 right parenthesis to the power of straight n
          equals straight C presuperscript straight n subscript 0 5 to the power of 0 plus straight C presuperscript straight n subscript 1 5 to the power of 1 plus straight C presuperscript straight n subscript 2 5 squared plus straight C presuperscript straight n subscript 3 5 cubed plus....... plus straight C presuperscript straight n subscript straight n 5 to the power of straight n
    rightwards double arrow                6 to the power of straight n space equals space 1 plus 5 straight n plus 5 squared left square bracket straight C presuperscript straight n subscript 2 plus 5 space straight C presuperscript straight n subscript 3 plus...... plus 5 to the power of straight n minus 2 end exponent right square bracket
    rightwards double arrow                space 6 to the power of straight n equals 1 plus 5 straight n plus 25 straight k
    where straight k equals straight C presuperscript straight n subscript 2 plus 5 space straight C presuperscript straight n subscript 3 plus........ plus 5 to the power of straight n minus 2 end exponent
    rightwards double arrow                     space space 6 to the power of straight n minus 5 straight n space equals space 25 straight k plus 1
    Hence,    6 to the power of straight n minus 5 straight n leaves a remainder 1 when divided by 25.
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    16.

    Find the value of space space open parentheses straight a squared plus square root of straight a squared minus 1 end root close parentheses to the power of 4 plus open parentheses straight a squared minus square root of straight a squared minus 1 end root close parentheses to the power of 4

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    17. Find the expansion of (3x2 - 2ax + 3a2)3 using binomial theorem. 
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    18. If a and b are distinct integers, prove that a-b is a factor of an - bn, whenever n is positive integer.
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    19. Find co-efficient of x5 in the expansion of (1 + 3x)6 (1 - x)5
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    20. In the expansion of (x + a)n, sums of odd and even terms are P and Q respectively. 
    Prove that:

    (a) 2 left parenthesis straight P squared plus straight Q squared right parenthesis space equals space left parenthesis straight x plus straight a right parenthesis to the power of 2 straight n end exponent plus left parenthesis straight x minus straight a right parenthesis to the power of 2 straight n end exponent  (b) left parenthesis straight P squared minus straight Q squared right parenthesis space equals space left parenthesis straight x squared minus straight a squared right parenthesis to the power of straight n
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