Let t_k denote the k^th term of the given series.
Then \(t_k{1\over k(k+1)}.\) 
By using partial fraction we get
\({1 \over k(k+1)}={1\over k}-{1\over{k+1}}\)
Thus \(t_1+t_2+....+t_n=\left( 1-{1\over 2} \right)+\left( {1\over 2}+{1\over 3} \right)+\left( {1\over 3}-{1\over 4} \right)+...+\left( {1\over n} -{1\over n+1} \right)=1-{1\over n+1}.\)

In (a+b)ⁿ, the general terms is T_r+1 = nC_r a^n-r b^r
To find the Coefficient of 4th term , put r = 3 in (1)
∴ T₄  = nC₃ a^n-3 b³
To find the Coefficient of 13^th them
Put r = 12 in (1)
T₁₃ = nC₁₂ a^n-12 b¹²
Given nC³ = nC₁₂
3 +12  = n
n = 15

Consider 3⁶⁰⁰
= (3²)³⁰⁰ = 9³⁰⁰
3⁶⁰⁰ = (10 -1 )³⁰⁰
Using binomial theorem
3⁶⁰⁰ = 300C₀ (10)³⁰⁰ - 300C₁(10)²⁹⁹ + ...-300C₂₉₉ (10)¹ + 1
= (10)³⁰⁰ - 300 (10)²⁹⁹ + ... - 300(10)+1
3⁶⁰⁰ = (10)³⁰⁰ - 300 (10)²⁹⁹ +... - 3000 + 1
Hence, it is clear that the last two digits in 3⁶⁰⁰ are 01

= [(x-a)ⁿ = xⁿ + ⁿC₁x^n-1(-a)¹+ⁿC₂x^n-1(-a)²+.....(-a)ⁿ]
= \(\left[ \left( { 2x }^{ 2 } \right) ^{ 4 }-4C_{ 1 }\left( 2x^{ 2 } \right) \left( 3\sqrt { 1-{ x }^{ 2 } } \right) ^{ 2 }+4C_{ 2 }\left( { 2x }^{ 2 } \right) (3\sqrt { 1-{ x }^{ 2 }) } ^{ 2 }-4C_{ 3 }({ 2x }^{ 2 })^{ 1 }(3\sqrt { 1-{ x }^{ 2 } } )^{ 3 }+(3\sqrt { 1-{ x }^{ 2 } } )^{ 4 } \right] \)  \(=\left[ \left( { 2x }^{ 2 } \right) ^{ 4 }-4C_{ 1 }\left( 2x^{ 2 } \right) ^{ 3 }\left( 3\sqrt { 1-{ x }^{ 2 } } \right) ^{ 1 }+4C_{ 2 }\left( { 2x }^{ 2 } \right) (3\sqrt { 1-{ x }^{ 2 }) } ^{ 2 }+4C_{ 3 }({ 2x }^{ 2 })(3\sqrt { 1-{ x }^{ 2 } } )^{ 3 }+(3\sqrt { 1-{ x }^{ 2 }) } ^{ 4 } \right] \)= \(2\left[ \left( { 2x }^{ 2 } \right) ^{ 4 }+4C\left( { 2x }^{ 2 } \right) ^{ 2 }(3\sqrt { 1-{ x }^{ 2 } } )^{ 2 }+(3\sqrt { 1-{ x }^{ 2 } } )^{ 4 } \right] \)
= \(2\left[ \left( 16{ x }^{ 8 } \right) +\frac { 4\times 3 }{ 2\times 1 } \times { 4x }^{ 4 }\times 9(1-{ x }^{ 2 })+{ 3 }^{ 4 }(1-{ x }^{ 2 })^{ 2 } \right] \)
= \(2\left[ 16{ x }^{ 8 }+216{ x }^{ 4 }(1-{ x }^{ 2 })+81(1-{ x }^{ 2 })^{ 2 } \right] \)

\(\sqrt [ 3 ]{ 126 } ={ (125) }^{ 1/3 }=(125+1)^{ 1/3 }=\left\{ 125\left( 1+\frac { 1 }{ 125 } \right) \right\} ^{ 1/3 }=(125)^{ 1/3 }\left[ 1+\frac { 1 }{ 125 } \right] ^{ 1/3 }\)
\(=5\left[ 1+\frac { 1 }{ 3 } \times \frac { 1 }{ 125 } +... \right] \left( \therefore \frac { 1 }{ 125 } <1 \right) =5\left[ 1+\frac { 1 }{ 3 } (0.008) \right] =5(1+0.002666)=5.01\)

The general term in the expansion of (1 + x)^m is T_r+1 = ⁿC_r(1)^m-r x^r
On putting r = 2, we get T₃ = ^mC₂(1)^m-2 x² =  ^mC₂ x²
∴ Coefficient of x² = ^mC₂
Also, coefficient of x² in the expansion of (1+x)^m is 6
∴ ^mC₂ = 6 ⇒ \(\frac{m(m-1)}{2.1}=6 \Rightarrow m(m-1)=12\)
⇒ m(m-1) = 4.3
⇒ m = 4 

T₈ and T₉

As n = 7 which is odd, the terms containing x⁴y³ and x³y⁴ are the two middle terns.
They are ⁷C₃ x⁴y³ and ⁷C₄x³y⁴ which are equal 35x⁴y³ and 35x³y⁴.

Here n = 6, which is even.
Thus the middle term in the expansion of (x +y)⁶ is the term containing \({x}^{{6\over 2}}{y}^{{6\over 2}},\) that is the term ⁶C₃ x³y³ which is equal to 20x³y³.

Given series is 3 - 6 + 9 - 12+ ...
= 3 (1) + 6 (- 1) + 9 (-1)² + 1² (- 1)³+ . . .
This is an arithmetic geometric (AG) series with correspondingA.P 3, 6, 9, 12 ... and G.P 1, -1, (-1)²,(-1)³.
\(\therefore\). n^th term of the given A. G. series is
= (n^th term of 3, 6, 9, ... ) (n^th term of 1, - 1, (-1)², ... )
= [3 + (n - 1)3] [1 (-1)^n-1]      [\(\because\) For AP, a = 3, d = 3 for GP = a = 1, r = -1]
= (3 + 3n - 3) (-1)^n-1
= 3n (-1)^n-l.

Since H is the H. M. between a and b,
\(we\quad get\quad H=\frac { 2ab }{ a+b } \quad ........(1)\)
\(LHS=(H-2a)(H-2b)\)
\(=\left( \frac { 2ab }{ a+b } -2a \right) \left( \frac { 2ab }{ a+b } -2b \right) \)
\(=\left( \frac { 2ab-2{ a }^{ 2 }-2ab }{ a+b } \right) \left( \frac { 2ab--2ab-{ ab }^{ 2 } }{ a+b } \right) \)
\(=\left( \frac { { -2a }^{ 2 } }{ a+b } \right) \left( \frac { { -2b }^{ 2 } }{ a+b } \right) =\left( \frac { { 4a }^{ 2 }{ b }^{ 2 } }{ { \left( a+b \right) }^{ 2 } } \right) { \left( \frac { 2ab }{ a+b } \right) }^{ 2 }\)
\(={ H }^{ 2 }[using\quad 1]\)

In (1+2x)⁸ , we have n = 8, x = 1, a = 2x.
T_r+1 = nC_r x^n-r a^r
⇒ T_r+1 = 8Cr(1)^8-r.(2x)^r = 8Cr 2^r.x^r  --- (1)
and T_r-1 = 8C_r-1.(2x)^r-1 = 8C_r-1.2^r-1x^r-1 ----- (2)
Dividing (2) ÷ (1) we get,
\(\frac{T_{r+1}}{T_r}=\frac{8C_r.2^{r}.x^{r}}{8C_{r-1}.2^{r-1}.x_{r-1}}=\frac{8!}{r!(8-r)!}.\frac{(r-1)!(8-r+1)}{8!}2x\)
= \(\frac{8-r+1}{r}.2r=\frac{9-r}{r}.2(2)\)   [since x=2]
= \(\frac{36-4r}{r}\)
Now T_r+1 ≥T _r if \(\frac{T_{r+1}}{T_{r}}\ge1\)
⇒ \(\frac{36-4r}{r}\ge1\)
⇒ 36 - 4r≥1
⇒ 5r ≤ 36
⇒ r ≤ \(\frac{36}{5}\)
⇒ r ≤ 7.2
∴ the greatest possible value of r is 7.
The greatest possible value of r is 7.

Given \({ \left( x-\frac { 1 }{ 2y } \right) }^{ 10 }\)
Here n = 10, x = x and \(a=\left( \frac { -1 }{ 2y } \right) \)
Middle term = \({ T }_{ \frac { 10+2 }{ 2 } }={ T }_{ 6 }\)
General term is \({ T }_{ r+1 }=nCr{ x }^{ n-r }{ a }^{ r }\)
Putting r = 5 we get,
\({ T }_{ 6 }=10{ C }_{ 5 }{ x }^{ 10-5 }{ \left[ -\frac { 1 }{ 2y } \right] }^{ 5 }=\frac { 10\times 9\times 8\times 7\times 6 }{ 5\times 4\times 3\times 2\times 1 } .{ x }^{ 5 }\left( \frac { -1 }{ 32.{ y }^{ 5 } } \right) \)
\(=-225.{ x }^{ 5 }.\frac { 1 }{ 32{ y }^{ 5 } } { T }_{ 6 }=\frac { -63{ x }^{ 5 } }{ 8{ y }^{ 5 } } \)

T_n = a + (n - 1)d
T_m+n = a + (m + n - 1)d
& T_m-n = a + (m - n - 1)d
T_m+n + T_m-n = a + (m + n - 1)d + a + (m - n - 1)d
= 2a + d(m + n - 1 + m - n - 1)
= 2a + d(2m - 2)
= 2[a + (m - 1)d]
T_m+n + T_m-n = 2. T_m

Consider (1.01)^1000000 - 10, 000
= ( 1 + 0.1 )^1000000 - 10000
= 100000 C₀ + 1000000 C₁ (O1) + 1000000 C₂ (.01)² + ...+ (0.1)^1000000  - 10, 000
= ( 1 + 1000000 x (0.1) + other postive terms ) -10000
 1 + other positive terms
(1.01)^1000000 - 10, 0000 > 0
1.01 ^1000000 > 10, 000
(1.01)^1000000  is larger