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Q1. If the ratio of the coefficients of 3rd and 4th terms in the expansion of is 1:2 then find the value of n.

Solution

Given expansion is    T subscript 3 over T subscript 4 equals fraction numerator n open parentheses n minus 1 close parentheses cross times 3 cross times 2 cross times 1 cross times 8 over denominator n open parentheses n minus 1 close parentheses open parentheses n minus 2 close parentheses cross times 2 cross times 1 cross times 4 end fraction equals 1 half rightwards double arrow n minus 2 equals 12 rightwards double arrow n equals 14
Q2. Find the  term independent of x  in the expansion of    :

Solution

We have:   The term will be independent of x  if the index of x is zero, i.e. 9-3r = 0, Thus, r = 3 Hence, term is independent of x and is given by 489888.
Q3. Show that  is divisible by 64, whenever n is a positive integer:

Solution

9 to the power of n plus 1 end exponent minus 8 n minus 9 equals 9 left parenthesis 1 plus 8 right parenthesis to the power of n minus 8 n minus 9 9 left parenthesis 1 plus 8 n plus C presuperscript n subscript 2 8 squared plus..... plus C presuperscript n subscript n 8 to the power of n right parenthesis minus 8 n minus 9 9 plus 72 n plus 64 x 9 left parenthesis C presuperscript n subscript 2 plus C presuperscript n subscript 3 8 plus C presuperscript n subscript 4 8 squared plus... plus C presuperscript n subscript n 8 to the power of n minus 2 end exponent right parenthesis minus 8 n minus 9 64 n plus 64 x 9 left parenthesis C presuperscript n subscript 2 plus C presuperscript n subscript 3 8 plus C presuperscript n subscript 4 8 squared plus... plus C presuperscript n subscript n 8 to the power of n minus 2 end exponent right parenthesis space w h i c h space i s space d i v i s i b l e space b y space 64 Hence,  is divisible by 64.
Q4. Find the general term in the expansion of  : 

Solution

The term of the expansion is given by:
Q5. Find the middle term in the expansion of .

Solution

As n is even, the middle term of the expansion   isterm. T subscript 6 equals T subscript 5 plus 1 end subscript equals C presuperscript 10 subscript 5 x to the power of 5 left parenthesis fraction numerator negative 1 over denominator 6 y end fraction right parenthesis to the power of 5 equals negative 7 over 216 x to the power of 5 over y to the power of 5
Q6. Find the coefficient of in the expansion of :

Solution

In the expansion of , Let  be the term containing . So, 12- r = 5 and r = 7 Hence, r = 7 Hence the coefficient of = 101376.
Q7. If the coefficients of   terms of are in arithmetic progression, then find the value of r:

Solution

  Since the terms of given expansion are in arithmetic progression, we have,             or (15-2r) (r +1) = (13-2r) (15 –r) or 15r + 15 - - 2r = 195 – 30r -13r + or - 56r + 180 = 0 or - 4r + 45 = 0 or (r - 5) (r- 9) = 0 r = 5, 9.
Q8. In the binomial expansion of ,the coefficients of the 5th, 6th and7h terms are in AP. Find all the values of n for which this can happen.

Solution

The coefficients of the 5th, 6th and 7th terms are Since the coefficients of the 5th, 6th and 7th terms are in AP, we have,
Q9. Using binomial theorem, evaluate .

Solution

We know that: 104 = 100 + 4


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