Clearly has \({ \left( 2x-\frac { 1 }{ { x }^{ 2 } } \right) }^{ 25 }\) 26 terms.
So, 11^th term from the end = (26 - 11 + 1)^th terms from the beginning.
=16^th term from the beginning = t₁₆
General term is t_{r + I} =25C_r (2x)^25-r \({ \left( \frac { -1 }{ { x }^{ 2 } } \right) }^{ r }\)
To get t₁₆, put r = 15
\(\therefore\)t₁₆=25C₁₅(2x)^25-15\(\times\)\({ \left( \frac { -1 }{ { x }^{ 2 } } \right) }^{ 15 }\)=25C₁₅2¹⁰\(\times\)x¹⁰\({ \left( \frac { -1 }{ { x }^{ 30 } } \right) }\)=-25C₁₅\(\frac { 2^{ 10 } }{ { x }^{ 20 } } \)

Given \(({\sqrt{2}+1})^{5}+{(\sqrt{2}-1)}^{5}\)
=[\({ \left( \sqrt { 2 } \right) }^{ 5 }\)+ 5C_I \({ \left( \sqrt { 2 } \right) }^{ 4}\) (1)¹ + 5C₂ \({ \left( \sqrt { 2 } \right) }^{ 3 }\) .(1)² + 5C₃ \({ \left( \sqrt { 2 } \right) }^{ 2 }\) . (1)³+ 5C₄ \({ \left( \sqrt { 2 } \right) }^{ 1}\) .(1)⁴ + (1)⁵] +[\({ \left( \sqrt { 2 } \right) }^{ 5 }\) - 5C_I \({ \left( \sqrt { 2 } \right) }^{ 4}\) (l)¹ + 5C₂ \({ \left( \sqrt { 2 } \right) }^{ 3 }\) (1)² - 5C₃ \({ \left( \sqrt { 2 } \right) }^{ 2 }\) (1)³+ 5C₄ \({ \left( \sqrt { 2 } \right) }^{ }\) (1)⁴ -1⁵ ]
=2[\({ \left( \sqrt { 2 } \right) }^{ 5 }\) + 10\({ \left( \sqrt { 2 } \right) }^{ 3 }\) +5 \({ \left( \sqrt { 2 } \right) }^{ }\)] = 2[ 4\(\sqrt { 2 } \) + 20\(\sqrt 2\) + 5.\(\sqrt 2\)] = 2[29.\( \sqrt { 2 } \)] = 58\( \sqrt 2\)

Let P (n)be the statement. 1.3+2.3²+3.3 + ... +n.3ⁿ = \({{(2n-1){3}^{n+1}+3}\over{4}} \) for all n \(\in\) N.
Step-1:
Put n = 1  \(\Rightarrow1.3(1){{(2-1){3}^{1+1}+3}\over{4}}={{3^2+3}\over{4}}={{12}\over{4}}\Rightarrow\ 3=3\)
\(\therefore\) P(1) is true.
Step-2:
Let us assume that P(k) is true
\(\therefore\)  1.3 + 2.3² + 3.3³ + ... + k.3k = \({{(2k-1){3}^{k+1}+3}\over{4}}\)  ....(1)
Step-3:
To prove that P (k + 1) is true i.e. to P.T. 1.3 + 2.3² + 3.3³ + ... + k.3^k + (k + 1)3^k+1
\(={{[2(k+1)-1]{3}^{k+2}+3}\over{4}}={{(2k+1){3}^{k+2}+3}\over{4}}\)
LHS = 1.3 + 2.3² + ... + k.3^k + (k+ 1)3^{k+ 1}
\(={{(2k-1){3}^{k+1}+3}\over{4}}+(k+1){3}^{k+1}={{(2k+1){3}^{k+1}+3+(4k+3){3}^{k+1}}\over{4}}\)
\(={{{3}^{k+1}(2k-1+4k+4)+3}\over{4}}={{{3}^{k+1}(6k+3)+3}\over{4}}={{{3}^{k+1}(2k+1)+3}\over{4}}={{{3}^{k+2}(2k+1)+3}\over{4}}\) = RHS
\(\therefore\)  P (k + 1) is true whenever P(k) is true.
\(\therefore\) By mathematical induction, p(n) is true for all values n.

In \({\left( ax+{{1}\over{x}} \right)}^{n},n=n,x=ax,a={{1}\over{x}}\)
Compare \(\left(ax+{{1}\over{x}} \right)^n\) with (x+a)ⁿ
\(\therefore\)  General term is t_r+1 = nC_rX^n-r a^r
\(\Rightarrow\)    t_r+1 = nCr (ax)^n-r \({\left( {{1}\over{x}} \right)}^{r}=nC_r{a}^{n-r}{x}^{n-2r}\)
To find the fourth term, put r = 3
\({t}_{4}=nC_3{a}^{n-3}{x}^{n-2(3)}={{5}\over{2}}\)
\(\Rightarrow\)    \(nC_3{a}^{n-3}{x}^{n-6}={{5}\over{2}}\)
Equating the powers of x both sides, we get
n - 6 = 0  \(\Rightarrow\)  n = 6
Putting n = 6 in (1) we get,
\(6{C}_{x}{a}^{6-3}{x}^{0}={{5}\over{2}}\)
\(\Rightarrow\)    \(6C_3a^3={{5}\over{2}}\)    \([\because\ n{C}_{r}=n{C}_{n-r}]\)
\(\Rightarrow\)                        
\(\Rightarrow\)   20 a³ = \({{5}\over{2}}\Rightarrow{a}^{3}={{5}\over{2\times20}}\)
\(\Rightarrow\)  \({a}^{3}={{1}\over{8}}={\left({{1}\over{2}} \right)}^{3}\)  
\(\Rightarrow\)   \(a={{1}\over{2}}\)
 \(\therefore\) n = 6 and \(a={{1}\over{2}}\)

A parallelogram is formed by choosing two straight lines from the set of m parallel lines and two straight lines from the set of n parallel lines. Two straight lines from the set of m parallel lines can be chosen in mC₂ ways and Two straight lines from the set of n parallel lines can be chosen in nC₂ ways
Hence, the number of parallelograms formed\(\Rightarrow\)mC₂\(\times \)nC₂\(=\frac { m(m-1) }{ 2\times 1 } \times \frac { n(n-1) }{ 2\times 1 } =\frac { mn(m-1)(n-1) }{ 4 } \)