Step I: Let P(n) be the given statement
i.e. P(n): \(1\times 1!+2\times 2!+3\times 3!+....+n\times n!=(n+1)!-1\)
Step II : For n=1, we have
LHS=\(1\times 1!\)=1
and RHS=(1+1)!-1=2!-1=2-1=LHS
\(\because \)  LHS=RHS
\(\therefore \)  P(1) is true
Step III  Let us assume that P(n) is true for n=k
Then, we have
P(k): \(1\times 1!+2\times 2!+3\times 3!+....+k\times k!+(k+1)!-1\quad \quad ....(i)\)
Step IV  Now, we shall prove the statement for n=k+1. For this we have to show that
\(1\times 1!+2\times 2!+3\times 3!+....+k\times k!+(k+1)\times (k+1)!=(k+1+1)!-1\)
Then, LHS \(=1\times 1!+2\times 2!+3\times 3!+....+k\times k!+(k+1)\times (k+1)!\)
=(k+1)!-1+(k+1)!\(\times \)(k+1)      [from Eq.(1)]
=(k+1+1)(k+1)!-1=(k+2)(k+1)!-1
=(k+2)!-1   [\(\because \)n(n-1)!=n]
Thus, P(k+1) is true, whenever P(k) is true. Hence, by the principle of mathematical induction, P(n) is true for all natural numbers n.

Step I  Let P(n) be the given statement
i.e   P(n): logxⁿ =n log x
Step II    For n=1, we have
LHS=log x^-1 =log x
\(\because \)  LHS=RHS
\(\therefore \)  P(1) is true.
Step III Let us assume that P(k) is true,
i.e. P(k): log x^k =k log x ..(i)
Step IV  Now, we shall prove the statement for n=k+1. For this,  we have to show that
P(k+1): log x^k+1 =(k+1) log x
Then, LHS=Log x^x+1 =log (x^k.x)=log x^k +log x    [log (m.n)=log m+log n]
=k log x+log x     [from Eq.(1)]
=(k+1) log x=RHS
Thus, P(k+1) is true, whenever P(k) is true. Hence, by principle of mathematical induction, P(n) is true for all \(n \in N\)

Let P(n) be the given statement,
i.e. P(n) : (ab)ⁿ=aⁿ  b ⁿ
We note that P(n) is true For n=1, (ab)¹ =ab=a¹ b¹
Let P(k) be true, i.e.,
Then, we have (ab)^k =a^k b^k  ..(i)
We shall now prove that P(k + 1) is true whenever P(k) is true.
Now, we have
(ab)^{k + 1} = (ab)^k (ab)
=( a^k b^k ) (ab)       [using Eq.(i)]
 =(a^k .a¹) (b^k.b¹)=a^k+1.b^k+1 
Thus, P(K+1) is also true, whenever P(k) is true. Hence, by principle of mathematical induction, P(n) is true for all \(n \in N\)

Step I  Let P(n) be the given statement.
i.e. P(n): \(1+3+3^{ 2 }+.....+3^{ n-1 }=\frac { 3^{ n }-1 }{ 2 } \)
Step II  For n=1, we have
LHS=3^1-1=3⁰=1 and RHS= \(\frac { 3^{ 1 }-1 }{ 2 } =1\)
\(\therefore \)   LHS=RHS 
\(\therefore \)   P(1) is true.
Step III  Let us assume that P(n) is true for n=k
Then, we have
\(P(k) :1+3+3^{ 2 }+.....+3^{ k-1 }=\frac { 3^{ k }-1 }{ 2 } ....(i)\)    
Step IV  Now, we shall prove the statement for n=k+1 For this,                  
we have to show
\(1+3+3^{ 2 }+.....+3^{ k-1 }+3^{ k }=\frac { 3^{ k+1 }-1 }{ 2 } \)
Then, LHS=1+3+3²+...3^k-1+3^k
\(=\frac { 3^{ k }-1 }{ 2 } +{ 3 }^{ k }\)      [from Eq. (i)]
\(=\frac { 3^{ k }-1+2.3^{ k } }{ 2 } =\frac { 3^{ k }(1+2)-1 }{ 2 } \)
\(=\frac { 3^{ k }.3-1 }{ 2 } =\frac { 3^{ k+1 }-1 }{ 2 } =RHS\)
Thus, P(k+1) is true, whenever P(k) is true. Hence, by principle of mathematical induction, P(n) is true for all \(n \in N\)

Step I  Let P(n) be the given statement,
i.e. P(n): \({ 1 }^{ 3 }+{ 2 }^{ 3 }+{ 3 }^{ 3 }+...+{ n }^{ 3 }=\left[ \frac { n(n+1) }{ 2 } \right] ^{ 2 }\)
Step II  For n=1, we have, LHS=1³=1
and RHS= \(\left[ \frac { 1(1+1) }{ 2 } \right] ^{ 3 }=\left( \frac { 1.2 }{ 2 } \right) ^{ 3 }=1\)
LHS=RHS;     P(1) is true
Step III  Let us assume that P(n) is true for n=k. Then we have
P(k): \({ 1 }^{ 3 }+{ 2 }^{ 3 }+{ 3 }^{ 3 }+.....+{ k }^{ 3 }=\left[ \frac { k(k+1) }{ 2 } \right] ^{ 2 }...(i)\)
Step IV  Now, we shall prove the statement for n=k+1 For this we have to show
1³+2³+3³+.....+k³+(k+1)³
=\(\left[ \frac { (k+1)(k+1+1) }{ 2 } \right] ^{ 2 }\)
Now, LHS=1³+2³+3³+.....+k³+(k+1)³
= \(\left[ \frac { k(k+1) }{ 2 } \right] ^{ 2 }+(k+1)^{ 3 }\)       [from Eq.(i)]
\(=(k+1)^{ 2 }\left[ \frac { k^{ 2 } }{ 4 } +(k+1) \right] \)
\(\\ =(k+1)^{ 2 }\left[ \frac { k^{ 2 }+4(k+1) }{ 4 } \right] =\frac { (k+1)^{ 2 }\left[ k^{ 2 }+4k+4 \right] }{ 4 } \)
\(=\frac { (k+1)^{ 2 }(k+2)^{ 2 } }{ 4 } =RHS\)
So, P(k+1) is true, whenever P(k) is true. Hence, by principle of mathematical induction, P(n) is true for all \(n\in N\)

0

i

\(\left( \frac { 1 }{ 2 } +\frac { 5 }{ 2 } i \right) -\frac { 3 }{ 2 } i+\left( -\frac { 5 }{ 2 } -i \right) \)
\(=\left( \frac { 1 }{ 2 } -\frac { 5 }{ 2 } \right) +i\left( \frac { 5 }{ 2 } -\frac { 3 }{ 2 } -1 \right) \)   
 = -2 + i0, which is in the form of a + ib.

\(z=\sqrt { 37 } +\sqrt { -19 } =\sqrt { 37 } +\sqrt { 19 } \sqrt { -1 } \)
\(z=\sqrt { 37 } +\sqrt { 19i } \quad [\because \sqrt { -1 } =i]\)
\(\therefore \ Re(z)=\sqrt { 37 } and \ Im(z)=\sqrt { 19 } \)

Consider :\(P(k) : \sum _{ t=1 }^{ k-1 }{ t(t+1) } =\frac { k(k-1)(k+1) }{ 3 } ​​​​k\ge 2\) 
P(k):1.2+2.3+3.4+....+(k-1)k \(=\frac { k(k-1)(k+1) }{ 3 } \)
Now P(k+1):1.2+2.3+3.4+...+(k1)k+k(k+1)
\(=\frac { k(k-1)(k+1) }{ 3 } +k(k+1)\)
\(=\frac { k(k+1)(k+2) }{ 3 } k\ge 2\)

Consider  P(k):2^2k -1= \(3\lambda \) (say)
Now P(k+1):2^2k+1 -1 = 2^2k.2² -1
\((3\lambda +1)4-1=12\lambda +3\)
\(3(4\lambda +1)\), which is divisible by 3.

\(\frac { 1 }{ 2.5 } +\frac { 1 }{ 5.8 } +\frac { 1 }{ 8.11 } +....+\frac { 1 }{ (3k-1)(3k+2) } =\frac { k }{ 6k+4 } \)
Now P(k+1):\(\frac { 1 }{ 2.5 } +\frac { 1 }{ 5.8 } +\frac { 1 }{ 8.11 } +....+\frac { 1 }{ (3k-1)(3k+2) } +\frac { 1 }{ [3(k+1)-1][3(k+1)+2] } \)
=\(\frac { k }{ 2(3k+2) } +\frac { 1 }{ (3k+2)(3k+5) } \)
=\(\frac { { 3k }^{ 2 }+5k+2 }{ 2(3k+2)(3k+5) } +\frac { (3k+2)(k+1) }{ 2(3k+2)(3k+5) } =\frac { k+1 }{ 2(3k+5) } \)
=\(\frac { k+1 }{ 6(k+1)+4 } \)

Consider 
P(k):\(\left( 1+\frac { 1 }{ 1 } \right) \left( 1+\frac { 1 }{ 2 } \right) \left( 1+\frac { 1 }{ 3 } \right) .....\left( 1+\frac { 1 }{ k } \right) =k+1\)
P(k+1):\(\left[ \left( 1+\frac { 1 }{ 1 } \right) \left( 1+\frac { 1 }{ 2 } \right) \left( 1+\frac { 1 }{ 3 } \right) .....\left( 1+\frac { 1 }{ k } \right) \right] \left( 1+\frac { 1 }{ k+1 } \right) \)
\(=(k+1)\left(\frac{k+2}{k+1}\right) \)
=(k+1)+1

Consider P(k) : 1+2+2²+.... +2^k = 2^k+1-1
Now P(+1):1+2+2² +..+2^k = 2^k+1 
=2^k+1-1+2^k+1
=2^(k+1)+1 -1

Consider P(k):2+2.3+2.3² + ....+2.3^k-1 = (3^k-1)
Now,P(k+1):2+2.3+2.3² + ....+2.3^k-1 +2.3^k
=(3^K -1)+2.3^k =3.3^k -1 =3^k+1-1

Consider P(k):2+4+6+8+.....+2k = k(k+1)
Now, P(k+1):2+4+6+.....2(k)+2(k+1)
=k(k+1)+2(k+1)=k² + 3k+2
=(k+1)(k+1)

Consider P(k):2+4+6+8+.....+2k = k(k+1)
Now, P(k+1):2+4+6+.....2(k)+2(k+1)
=k(k+1)+2(k+1)=k² + 3k+2
=(k+1)(k+1)

We have, a + ib = \(\frac { ({ x }^{ 2 }+1) }{ 2{ x }^{ 2 }+1 } \) ....(i)
Take modulus both sides of E_q . (i) and then solve it.

We have, 
\(x+i y=\frac{(a+i b)(a+i b)}{(a-i b)(a+i b)}=\frac{a^{2}-b^{2}+2 a b i}{a^{2}+b^{2}}=\frac{a^{2}-b^{2}}{a^{2}+b^{2}}+\frac{2 a b}{a^{2}+b^{2}} i\)
So that, \(x-i y=\frac{a^{2}-b^{2}}{a^{2}+b^{2}}-\frac{2 a b}{a^{2}+b^{2}} i\)
Therefore,
\(x^{2}+y^{2}=(x+i y)(x-i y)=\frac{\left(a^{2}-b^{2}\right)^{2}}{\left(a^{2}+b^{2}\right)^{2}}+\frac{4 a^{2} b^{2}}{\left(a^{2}+b^{2}\right)^{2}}=\frac{\left(a^{2}+b^{2}\right)^{2}}{\left(a^{2}+b^{2}\right)^{2}}=1\)

Here z=1-i=r(cos \(\theta \)+i sin \(\theta \))
⇒  r cos \(\theta \) =1 and r sin \(\theta \) = -1...(i)
r²(cos²\(\theta \)+sin²\(\theta \))=1+1
⇒ \(\sqrt { 2 } \)cos\(\theta \) = 1 and \(\sqrt { 2 } \)sin\(\theta \) =1
⇒ cos\(\theta \)=\(-\frac { 1 }{ \sqrt { 2 } } \) and sin\(\theta \) = \(\frac { 1 }{ \sqrt { 2 } } \)
Since sin\(\theta \) is negative and cosf is positive
∴ \(\theta \) lies in fourth quadrant
∴   \(\theta =\left( \pi -\frac { \pi }{ 4 } \right) =\frac { 3\pi }{ 4 } \)
Hence polar form of z is \(\sqrt { 2 } \left[ cos\left( \frac { -\pi }{ 4 } \right) +i\quad sin\left( \frac { -\pi }{ 4 } \right) \right] \)

Since, -3 + ix²y and x² + y + 4i are conjugate of each other \(-3+i{ x }^{ 2 }y=\overline { { x }^{ 2 }+y+4i } x = {-b \pm \sqrt{b^2-4ac} \over 2a}\).
After this, equate real and imaginary parts, to get the values of x and y.  (x = 1, y = -4) or (x = -1, y = -4)

We have, \(\frac{(3-2 i)(2+3 i)}{(1+2 i)(2-i)}\)
\( =\frac{6+9 i-4 i+6}{2-i+4 i+2}=\frac{12+5 i}{4+3 i} \times \frac{4-3 i}{4-3 i}\)
\( =\frac{48-36 i+20 i+15}{16+9}=\frac{63-16 i}{25}=\frac{63}{25}-\frac{16}{25} i\)
Therefore, conjugate of \(\frac{(3-2 i)(2+3 i)}{(1+2 i)(2-i)}\) is \(\frac{63}{25}+\frac{16}{25} i\).

\(\left( -i \right) \left( 3i \right) { \left( -\frac { 1 }{ 6 } i \right) }^{ 3 }=\quad \left( -3i \right) \left( -\frac { 1 }{ 216 } { i }^{ 3 } \right) \)
\(=\left( -3\times \left( -1 \right) \right) \left( -\frac { 1 }{ 216 } \left( -i \right) \right) \quad \left[ \because { i }^{ 2 }=-1\quad and\quad { i }^{ 3 }=-i \right] \)
\(=3\times \frac { 1 }{ 216 } \times i\)
\(=\frac { i }{ 72 } =0+\frac { 1 }{ 72 } i\) 
Which is in form of a + ib.

We have, (x² + 2xi) - (3x² + iy ) = (3 - 5i) + (1 + 2iy)
\(\Rightarrow \) (x⁴ - 3x²) + (2x - y)i = 4 + (-5 + 2y)i
On equating real and imaginary parts both sides, we get
x⁴ - 3x² = 4 ...(i)
and  2x - y = -5 + 2y
 \(\Rightarrow \)  2x - 3y = -5......(ii)
On solving Eq. (i), we get
x⁴ - 3x² = 4 \(\Rightarrow \) x⁴ - 3x² - 4 = 0
\(\Rightarrow \)  x⁴ - 4x² + x² - 4 = 0
\(\Rightarrow \) (x² - 4)(x² + 1) = 0 \(\Rightarrow \) x² - 4 = 0
[\(\therefore \) x² + 1 \(\neq \) 0, for any real value of x]
\(\therefore \)  x =\(\pm \)2
On putting x =\(\pm \)2 in Eq. (ii), we get
y = 3, when x =2 and y = \(\frac { 1 }{ 3 } \), when x = -2
Thus, x = -2, y = \(\frac { 1 }{ 3 } \) or x = 2, y = 3.

We have, \(\sqrt { -25 } +3\sqrt { -4 } +2\sqrt { -9 } \)
\(=\sqrt { 25 } \sqrt { -1 } +3\sqrt { 4 } \sqrt { -1 } +2\sqrt { 9 } \sqrt { -1 }\)
= 5 x i + 3 x 2 x i + 2 x 3 i
= 5i + 6i + 6i
= 17i