Let P(n) denote the statement
1 + 3 + 5 +…+ (2n – 1) = n²
Put n = 1
LHS = 1
RHS = 1² = 1
LHS = RHS `
\(\therefore\) P(1) is true.
Assume that P(k) is true.
i.e., 1 + 3 + 5 + ... + (2k – 1) = k<sup>2</sup>
To prove: P(k + 1) is true. `
\(\therefore\) 1 + 3 + 5 + … + (2k – 1) + (2k + 1)
\(=p(k)+(2k+1)\)
\(={ k }^{ 2 }+2k+1\)
\(=\left( k+1 \right) ^{ 2 }\)
P(k + 1) is true whenever P(k) is true. `
\(\therefore\) P(n) is true for all \(n \in N\)

A Cricket team of 11 players can be formed in the following ways:
(i) 3 bowlers, 1 wicket keeper and 7 other players can be selected in 
\({ 4C }_{ 3 }\times { 2C }_{ 1 }\times { 10C }_{ 7 }\)  ways
\(4{ C }_{ 3 }\times { 2C }_{ 1 }\times { 10C }_{ 7 }={ 4C }_{ 1 }\times { 2C }_{ 1 }\times { 10C }_{ 3 }\) 
= 960 ways 
(ii) 3 bowlers, 2 wicket keepers and 6 other players can be selected in
\(\\ { 4C }_{ 1 }\times 2{ C }_{ 2 }\times { 10C }_{ 6 }\)  ways
\({ 4C }_{ 1 }\times { 2C }_{ 2 }\times { 10C }_{ 6 }=4{ C }_{ 1 }\times { 2C }_{ 2 }\times { 10C }_{ 4 }\) 
= 840 ways 
(iii) 4 bowlers and 1 wicket keeper and 6 other players 
\({ 4C }_{ 4 }\times { 2C }_{ 1 }\times 10{ C }_{ 6 }\)  ways
\(= { 4C }_{ 4 }\times { 2C }_{ 4 }\times { 10C }_{ 4 }=420\ ways\) 
4 bowlers, 2 wicket keepers and 5 other players can be selected in
\({ 4C }_{ 4 }\times { 2C }_{ 2 }\times { 10C }\) ways 
\({ 4C }_{ 4 }\times { 2C }_{ 2 }\times { 10C }_{ 5 }=252\) ways 
By addition principle of counting. Total number of ways
= 960 + 840 + 420 + 252
= 2472

Here x² + 1 cannot be factorized into linear factors.
Let \(\cfrac { 2x+1 }{ \left( x-1 \right) \left( { x }^{ 2 }+1 \right) } =\cfrac { A }{ x-1 } +\cfrac { Bx+C }{ { x }^{ 2 }+1 } \) ...(1)
Multiplying both sides by \((x-1)\left( { x }^{ 2 }+1 \right) \) 
\(2x+1=A\left( { x }^{ 2 }+1 \right) +\left( Bx+C \right) \left( x-1 \right) \) ..(2)
Put x =1 in (2)
\(2+1=A(1+1)+0\)
\(\therefore A=\cfrac { 3 }{ 2 } \) 
Put x = 0 in (2)
\(0+1=A(0+1)+(0+C)(-1)\)
\(1=A-C\)
\(\therefore C=\cfrac { 1 }{ 2 } \) 
Equating the coefficient of x² on both the sides of (2), we get
\(A+B=0\)
\(B=-A\)
\( \therefore B=\cfrac { -3 }{ 2 } \) 
Substituting the values of A, B and C in (1), we get
\(\cfrac { 2x+1 }{ (x-1)\left( { x }^{ 2 }+1 \right) } =\cfrac { \frac { 3 }{ 2 } }{ x-1 } +\cfrac { \frac { -3 }{ 2 } x+\frac { 1 }{ 2 } }{ { x }^{ 2 }+1 } \)
\(=\cfrac { 3 }{ 2(x-1) } -\cfrac { 3x-1 }{ 2({ x }^{ 2 }+1) } \)

\(\text {Let } \frac{x+1}{(x-2)^2(x+3)} =\frac{A}{(x-2)}+\frac{B}{(x-2)^2}+\frac{C}{(x+3)}\) ........(1)
Multiplying both the sides by (x - 2)²(x + 3), we get
x + 1 = A(x - 2)(x + 3) + B(x + 3) + C(x - 2)²...(2)
Put x = 2 in (2), we have
2 +1 = A(0) + B(5) + C(0)
\(\therefore B=\frac{3}{5}\)
Put x = –3 in (2), we have
–3 + 1 = A(0) + B(0) + C(-3 - 2)²
\(\therefore C=\frac{-2}{25}\)
Equating the coefficient of x² on both the sides of (2), we get
0 = A + C
A = -C
\(\therefore A=\frac{2}{25}\)
Substituting the values of A,B and C in (1),
\(\frac{x+1}{(x-2)^2(x+3)}= \frac{2}{25(x-2)}+\frac{3}{5(x-2)^2}-\frac{2}{25(x+3)}\)

Write the denominator into the product of linear factors
Here (x² - 4)(x + 1)=(x - 2)(x + 2)(x + 1)
\(\therefore \frac{x+4}{\left(x^2-4\right)(x+1)}= \frac{A}{(x-2)}+\frac{B}{(x+2)}+\frac{C}{(x+1)}\)   ...(1)
Multiplying both the sides by (x – 2)(x + 2)(x + 1), we get
x + 4 = A(x + 2)(x + 1)+ B(x - 2)(x + 1)+ C(x - 2)(x + 2)  .....(2)
Put x = –2 in (2), we get
-2 + 4 = A(0) + B(-4)(-1) + C(0)
\(\therefore B=\frac{1}{2}\)
Put x = 2 in (2), we get
2 + 4 = A(4)(3) + B(0) + C(0)​
\(\therefore\ A=\frac{1}{2}\)
Put x = –1 in (2), we get
-1 + 4 = A(0) + B(0) + C(-3)(1)​
\(\therefore\) C = -1
Substituting the values of A, B and C in (1), we get
\(\frac{x+4}{\left(x^2-4\right)(x+1)}=\frac{1}{2(x-2)}+\frac{1}{2(x+2)}-\frac{1}{x+1}\)