\({ a }_{ n }=\begin{cases} 1 \\ 2 \\ { a }_{ n-1 }+{ a }_{ n-2 }ifn>2 \end{cases}\)
a₁ = 1, a₂ = 2
a₃ = a₂ + a₁ =2 +1 = 3
a₄ =a₃ + a₂ = 3+2 = 5
a₅ = a₄ + a₃ = 5 +3 = 8
a₆ = a₅ + a₄ = 8 +5 = 13
hence the first 6 terms are 1, 2, 3, 5, 8, 13

Here a = 1 d = 5 and \(r={1\over 7}\)  \(S_n={a-(a+(n-1)d)r^n\over 1-r}+dr\left( {1-{r}^{n-1}\over(1-r)^2} \right)\)
\(={1-(1+5(n-1)){({1\over 7})}^{n}\over1-{1\over7}}+5\times{1\over 7}\left( {{1-\left( 1\over 7 \right)^{n-1}}\over{\left( 1-{1\over 7}\right)}^2 } \right)\)
\(={{1-{5n-4\over 7^n}}\over{6\over 7}}+{{5\over 7}({7}^{n-1}-1)\over{7}^{n-1}{\left( {6\over 7} \right)}^{2}}\Rightarrow {7^n-5n+4\over {7}^{n-1}6}+{5({7}^{n-1}-1)\over{7n}^{-1}6}\)

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.

By taking a = 100, b = 2 and n = 4 in the binomial expansion of (a - b)ⁿ we get
98⁴ = (100-2)⁴
= ⁴C₀100⁴ - ⁴C₁ 100³2 + ⁴C₂ 100²2²- ⁴C3 100¹2³ + ⁴C₄ 100⁰2⁴
= 100000000 - 8000000 + 240000 - 3200 + 16
= 92236816.

By taking a = 2x, b = 3 and n = 5 in the binomial expansion of (a + b)ⁿ we get
(2x + 3)⁵ = (2x)⁵ + 5(2x)⁴3 + 10(2x)³3² + 10(2x)²3³ + 5(2x)3⁴ + 3⁵
= 32x⁵ + 240x⁴ + 720x³ + 1080x² + 810x + 243.

odd terms are 6, 4, 2, 0...
t_n = 6 +( n -1 ) (-2) = 6 -2n + 2
= 8 -2n
Even terms are 10, 12, 14 , 16
Here a = 1 , d = 2
t_n = 10 + ( n - 1) (2) = 10 + 2n -2
= 8 + 2n
n^th term of the given sequence is  \(\begin{cases} 8-2n \\ 8+2n \end{cases}\)

Numerators are 1, 3, 4, 7, 9
a = 1 d = 2 -1 
a_n  = 1 + ( n -1) 2 = 1 + 2n - 2 = 2n -1
denominator  2, 4, 6, 8, 10
a = 2, d = 2
a_n = 1 +( n- 1) 2 = 1 + 2n - 2 = 2n
Hence n^th term of the given sequence is \(\frac { 2n-1 }{ 2n } =1-\frac { 1 }{ 2n } \)

Consider the terms in the numerator 1, 2, 3....
a = 1, d = 2 -1 = 1 a_n = a + (n-1) d
a_n = 1 + (n-1) (1) = 1 + n - 1 = n
The terms in the denominator are 2, 3, 4, 5, 6...
here  a = 2, d = 1
a_n = 2 + (n-1) 1 = 2 + n -1 = n + 1
Hence n^th term of the given sequence is \(\frac { n }{ n+1 } \)
 

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