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

LHS = \({ \left( { x }^{ 3 }+6 \right) }^{ \frac { 1 }{ 3 } }-{ \left( { x }^{ 3 }+3 \right) }^{ \frac { 1 }{ 3 } }\)
\(={ x }^{ 3\times \frac { 1 }{ 3 } }{ \left( 1+\frac { 6 }{ { x }^{ 3 } } \right) }^{ \frac { 1 }{ 3 } }-{ x }^{ 3\times \frac { 1 }{ 3 } }{ \left( 1+\frac { 3 }{ { x }^{ 3 } } \right) }^{ \frac { 1 }{ 3 } }\)
\(=x\left[ 1+\frac { 1 }{ 3 } \left( \frac { 6 }{ { x }^{ 3 } } \right) \right] -x\left[ 1+\frac { 1 }{ 3 } \left( \frac { 3 }{ { x }^{ 3 } } \right) \right] \)
\(=x+\frac { 2 }{ { x }^{ 2 } } -x-\frac { 1 }{ { x }^{ 2 } } \)
\(=\frac { 2 }{ { x }^{ 2 } } -\frac { 1 }{ { x }^{ 2 } } =\frac { 1 }{ { x }^{ 2 } } =RHS\)
Hence proved.

2,2,4,4,6,6
Given sequences is 2, 2, 4, 4, 6, 6,
the odd term are 2, 4, 6 .. and even terms are also 2, 4, 6
\(\therefore { \ a }_{ n= }\begin{cases} n+1 \\ 1 \end{cases}\)
if n is odd
if n is even

\(\frac { 1 }{ 2^{ n+1 } } \)
\({ a }_{ 1 }\frac { 1 }{ 2^{ n+1 } } =\frac { 1 }{ { 2 }^{ 2 } } ,{ a }_{ 2 }=\frac { 1 }{ { 2 }^{ 2+1 } } \frac { 1 }{ { 2 }^{ 3 } } \)
\({ a }_{ 2 }=\frac { 1 }{ 2^{ 3+1 } } =\frac { 1 }{ { 2 }^{ 4 } } \)
\({ a }_{ 3 }=\frac { 1 }{ 2^{ 4+1 } } =\frac { 1 }{ { 2 }^{ 5 } } ,{ a }_{ 5 }=\frac { 1 }{ 2^{ 5+1 } } =\frac { 1 }{ { 2 }^{ 6 } } ,{ a }_{ 6 }=\frac { 1 }{ { 2 }^{ 6+1 } } =\frac { 1 }{ { 2 }^{ 7 } } \)
ஃ  the first 6 terms of the sequence are \(\frac { 1 }{ { 2 }^{ 2 } } ,\frac { 1 }{ { 2 }^{ 3 } } ,\frac { 1 }{ { 2 }^{ 4 } } ,\frac { 1 }{ { 2 }^{ 5 } } ,\frac { 1 }{ { 2 }^{ 6 } } and\frac { 1 }{ { 2 }^{ 7 } } \)
Since \({ a }_{ 1 }=\frac { 1 }{ { 2 }^{ 2 } } \& r=\frac { 1 }{ { 2 }^{ 3 } } \div \frac { 1 }{ { 2 }^{ 2 } } =\frac { 1 }{ { 2 }^{ 3 } } \times { 2 }^{ 2 }=\frac { 1 }{ 2 } \)
\(r=\frac { 1 }{ { 2 }^{ 4 } } \div \frac { 1 }{ { 2 }^{ 3 } } =\frac { 1 }{ 2 } \times { 2 }^{ 3 }=\frac { 1 }{ 2 } \)
the given sequence is a geometric progression

If a, b, c are HP, then \({1\over a},{1\over b},{1\over c}\) are in AP.
Then we have \({2\over b}={1\over a}+{1\over c},\) which gives ab-ac = ac - bc.
So, a(b-c) = c(a-b), which gives \({a\over c}={a-b\over b-c}.\) 
On the other hand, if \({a\over c}={a-b\over b-c},\) then a(b-c) = c(a - b).
Dividing each term by abc, we get \({1\over c}-{1\over b}={1\over b}-{1\over a}.\) 
Thus \({1\over a},{1\over b},{1\over c}\) are in AP and hence a, b, c are in HP. 

Let us take a = 3 and b = 2x in the binomial expansion of (a + b)¹⁰.
Then, x⁶ will appear in the term containing (2x)⁶ and nowhere else. So the term containing x⁶ is
\(^{10}{C}_{4}a^4b^6={10\times 9\times8\times 7\over4\times 3\times 2\times 1 }3^4{(2x)}^{6}=210\times3^4\times2^6x^6\)
So coefficient of x³ in the expansion of (3 + 2x)¹⁰ is 210 \(\times\) 3⁴2⁶