Explanation
There are total of n brackets. The term xn−2 will be formed when integers are chosen from any two brackets and x is chosen from all the other brackets and multiplied.
Thus, the Coefficient of xn−2 is
C=(1×2+1×3+...+1×n)+(2×3+2×4+..+2×n)+...+((n−1)×n)
=((n)(n+1)2−1)+2((n)(n+1)2−1−2)+...+(n−1)((n)(n+1)2−(1+2+3+..+(n−1)))
={(1+2+3+...+(n−1))((n)(n+1)2)}−{1+2(1+2)+3(1+2+3)+...+(n−1)(1+...+n−1)}
={((n−1)(n)2)((n)(n+1)2)}−{n−1∑1k((k)(k+1)2)}
={n2(n2−1)4}−{n−1∑1k3+k22}
={n2(n2−1)4}−12{((n−1)(n)2)2+(n−1)n(2n−1)6} =n(n−1)4{n(n+1)−n(n−1)2−2n−13}
=n(n−1)4{n(n+3)2−2n−13}
=n(n−1)4{3n2+9n−4n+26}
=n(n−1)4{(3n+2)(n+1)6} ∴ Option B is correct.
Put y=3log3√9|x−2| ⇒ log3y=log3√9|x−2| ⇒y=√9|x−2|=3|x−2| Now, put z=7(15)log7[(4).3|x−2|−9] log7z=15log7[(4).3|x−2|−9] =(log7[(4).3|x−2|−9])15 ⇒z=[(4).3|x−2|−9]15 Now, E=(y+z)7 and 6th term is given by t6=7C5y7−5z5=21(3|x−2|)2{(4).3|x−2|−9} ⇒567=21(32|x−2|){(4).3|x−2|−9} ⇒27=[(4)33|x−2|]−[(9)(32|x−2|)] ⇒27=(4)33|x−2|−(9)32|x−2| ⇒4u3−9u2−27=0;u=3|x−2| note that u=3 satisfies this equation 3|x−2|=3⇒|x−2|=1⇒x−2=±1 x=2±1=3or1.
{ (1+x) }^{ n }=^{ n }C_{ 0 }+^{ n }C_{ 1 }x+^{ n }C_{ 2 }{ x }^{ 2 }.........^{ n }C_{ n }{ x }^{ n }\\ \Rightarrow ^{ n }C_{ 0 }+^{ n }C_{ 1 }+^{ n }C_{ 2 }.........^{ n }C_{ n }={ 2 }^{ n }
For the given expansion { (1+x) }^{ 59 } let sum of last 30 coefficients be S
S=^{ 59 }C_{ 30 }+^{ 59 }C_{ 31 }+^{ 59 }C_{ 32 }..........^{ 59 }C_{ 59 } ....(i)
As ^{ n }C_{ r }=^{ n }C_{ n-r }
\Rightarrow S=^{ 59 }C_{ 29 }+^{ 59 }C_{ 28 }+^{ 59 }C_{ 27 }..........^{ 59 }C_{ 0 } ....(ii)
Adding (i) and (ii), we get
\Rightarrow 2S=^{ 59 }C_{ 0 }+^{ 59 }C_{ 1 }+^{ 59 }C_{ 2 }........^{ 59 }C_{ 59 }\\ \Rightarrow 2S={ 2 }^{ 59 }\\ \Rightarrow S={ 2 }^{ 58 }
So, option B is correct.
Consider the given expression.
\Rightarrow {{\left( {{2}^{x}}+\dfrac{1}{{{4}^{x}}} \right)}^{n}}
Given:
\dfrac{{{t}_{3}}}{{{t}_{2}}}=7
By binomial expansion, we know that
{{t}_{3}}={}^{n}{{C}_{2}}{{\left( {{2}^{x}} \right)}^{n-2}}{{\left( \dfrac{1}{{{4}^{x}}} \right)}^{2}}
{{t}_{2}}={}^{n}{{C}_{1}}{{\left( {{2}^{x}} \right)}^{n-1}}{{\left( \dfrac{1}{{{4}^{x}}} \right)}^{1}}
Again, we have
{}^{n}{{C}_{1}}+{}^{n}{{C}_{2}}=36
{}^{n+1}{{C}_{2}}=36
\dfrac{\left( n+1 \right)!}{2!\left( n+1-2 \right)!}=36
\dfrac{n\left( n+1 \right)}{2}=36
{{n}^{2}}+n-72=0
n=8,-9
Therefore,
n=8
Now,
\dfrac{{{t}_{3}}}{{{t}_{2}}}=\dfrac{{}^{n}{{C}_{2}}{{\left( {{2}^{x}} \right)}^{n-2}}{{\left( \dfrac{1}{{{4}^{x}}} \right)}^{2}}}{{}^{n}{{C}_{1}}{{\left( {{2}^{x}} \right)}^{n-1}}{{\left( \dfrac{1}{{{4}^{x}}} \right)}^{1}}}
\dfrac{{}^{8}{{C}_{2}}{{\left( {{2}^{x}} \right)}^{8-2}}{{\left( \dfrac{1}{{{4}^{x}}} \right)}^{2}}}{{}^{8}{{C}_{1}}{{\left( {{2}^{x}} \right)}^{8-1}}{{\left( \dfrac{1}{{{4}^{x}}} \right)}^{1}}}=7
\dfrac{28{{\left( {{2}^{x}} \right)}^{6}}{{\left( \dfrac{1}{{{\left( {{2}^{2}} \right)}^{x}}} \right)}^{2}}}{8{{\left( {{2}^{x}} \right)}^{7}}{{\left( \dfrac{1}{{{\left( {{2}^{2}} \right)}^{x}}} \right)}^{1}}}=7
\dfrac{7\times {{2}^{6x-4x}}}{2\times {{2}^{7x-2x}}}=7
{{2}^{2x-5x}}=2
-3x=1
x=-\dfrac{1}{3}
Hence, this is the required result.
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