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Question

Consider the binomial expansion R=(1+2x)n=I+f, where I is the integral part of R and f is the fractional part of R for nN. If the sum of the coefficients of R is 6561 and K=n+RRf, then which of the following is/are correct?

A
K=9 when x=12
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B
K=7 when x=12
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C
5th term is the greatest term when x=12
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D
If kth term has the greatest coefficient, then the sum of all possible value(s) of k is 13
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Solution

The correct option is D If kth term has the greatest coefficient, then the sum of all possible value(s) of k is 13
R=(1+2x)n=I+f
Put x=1 to get sum of all the coefficients
3n=6561=38n=8

For x=12 ; R=(2+1)8
Now, consider
(2+1)8+(21)8I+f+f=2[ 8C0(2)8+....]=even integer

Since, I is integer f+f must be an integer.
But 0<f+f<2
f+f=1f=1f

Now, K=n+RRf
K=n+R(1f)=8+(2+1)n×(21)n
K=8+1=9

For R=(1+2x)8,
Tr+1=8Cr(2x)r
Tr+1=8Cr when x=12
Now, Tr+1Tr
Tr+1Tr1
8Cr 8Cr11
8!r!(8r)!×(r1)!(9r)!8!1
9rr1
92r
For r=1,2,3,4 this is true
i.e., T5>T4
But for r=5, T6<T5
T5 is the greatest term.


Again, Tk+1=8Ck×2k×xk;
Tk=8Ck1×2k1×xk1;
Tk1=8Ck2×2k2×xk2
We have to find the term having the greatest coefficient.
2k1×8Ck12k×8Ck ...(1)
And 2k1×8Ck12k2×8Ck2 ...(2)

From (1),
8!×2k1(k1)!(9k)!2k×8!k!(8k)!1(9k)2kk183kk6

Again, 2k1×8Ck12k2×8Ck2
8!×2k1(k1)!(9k)!2k2×8!(k2)!(10k)!2k1110k202kk1k76k7
T6 and T7 term have the greatest coefficient.
k=6 or 7
Sum =6+7=13

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