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=38⇒n=8
For x=1√2 ; R=(√2+1)8
Now, consider
(√2+1)8+(√2−1)8I+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′=1⇒f′=1−f
Now, K=n+R−Rf
⇒K=n+R(1−f)=8+(√2+1)n×(√2−1)n
⇒K=8+1=9
For R=(1+2x)8,
Tr+1=8Cr(2x)r
⇒Tr+1=8Cr when x=12
Now, Tr+1≥Tr
⇒Tr+1Tr≥1
⇒ 8Cr 8Cr−1≥1
⇒8!r!(8−r)!×(r−1)!(9−r)!8!≥1
⇒9−rr≥1
⇒9≥2r
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=8Ck−1×2k−1×xk−1;
Tk−1=8Ck−2×2k−2×xk−2
We have to find the term having the greatest coefficient.
∴2k−1×8Ck−1≥2k×8Ck ...(1)
And 2k−1×8Ck−1≥2k−2×8Ck−2 ...(2)
From (1),
8!×2k−1(k−1)!(9−k)!≥2k×8!k!(8−k)!⇒1(9−k)≥2k⇒k≥18−3k⇒k≥6
Again, 2k−1×8Ck−1≥2k−2×8Ck−2
⇒8!×2k−1(k−1)!(9−k)!≥2k−2×8!(k−2)!(10−k)!⇒2k−1≥110−k⇒20−2k≥k−1⇒k≤7⇒6≤k≤7
⇒T6 and T7 term have the greatest coefficient.
⇒k=6 or 7
∴ Sum =6+7=13