Test: Special Sequences And Series

Given that (1.3)², (2.5)² , (3.7)² .....................20 terms
Here 1,2,3,.................are in A.P.
a = 1, d = 1
t_n = a + (n-1) d = 1 + (n -1)1 = n
3,5,7 ............. are in A.P
a = 3, d = 2
tn = a + (n-1) d = 3 + (n -1)² = 2n + 1
∴ nth term = n(2n + 1)² = 4n³ + 4n² + n
Sum of n terms Sn = ∑ ( 4n³ + 4n² + n ) = 4 { n²(n+1)²}/4 + 4 {n (n+1)(2n+1)} / 6 + n(n+1)/2
= { n²(n+1)²} + 2 {n (n+1)(2n+1)} / 3 + n(n+1)/2
Sum of 20 terms S₂₀ = 400 × 441 + 40 × 41 × 7 + 10 ×21 = 176400 + 11480 + 210
 = 188090.

0.2 + 0.22 + 0.222 + 0.2222 + ….
2(0.1 + 0.11 + 0.111 + 0.1111 + ….)
2/9 (0.9 + 0.99 + 0.999 + 0.9999 + ….)
2/9 (( 1 - 0.1 ) + ( 1 - 0.01 ) + ( 1 - 0.001 ) + ( 1 - 0.0001 ) + …. )
2/9 (1 - 0.1 + 1 - 0.01 + 1 - 0.001 + 1 - 0.0001 + ….)
2/9 ((1 + 1 + 1 + 1 + ….) - ( 0.1 + 0.01 + 0.001 + 0.0001 + ….))
2/9 (n - ( 0.1 + 0.01 + 0.001 + 0.0001 + ….))
2/9 (n - (( 0.1 ( 1 - 0.1ⁿ)/( 1 - 0.1))))
2/9 (n - (( 0.1 / 0.9)( 1 - 0.1ⁿ)))
2/9 (n - (1/9)(1 - 0.1ⁿ))

3² + 6² +....
∑(3n)²
9∑n²
9n(n+1)(2n+1)/6
= 3n(n+1)(2n+1)/2

The given word consists of 7 different letters out of which 3 are vowels and 4 are consonants.
7 different letters can be arranged in 
7! ways = 5040 ways.

S_n = 1/6n(n+1)(2n+1)
Here n = 15 so,
S₁₅ = 1/6 × 15(15+1)(2×15+1)
S₁₅ = 1/6 × 15 × 16(30 + 1)
S₁₅ = 1/6 × 15 × 16 × 31
S₁₅ = 1240

The formula to find sum of first n terms of a sequence of squares of natural numbers = n(n+1)(2n+1)/6

1³ + 3³ + 5³ + 7³ + 9³ − (2³ + 4³ + 6³ + 8³)
= Sum of cubes of first 5 odd number -Sum of cubes of first 4 even number
= n²(2n²–1) − 2n²(n+1)²
= 1225 – 800
= 425

S_n = 2 + [2+3¹] + [2+3¹+3²] + ⋯ + [2+3¹+3²+⋯+3^n-1] 
 = 2n+[3¹]+[3¹+3²]+⋯+[3¹+3²+⋯+3^n-1]
 = 2n + t_3,1 + t_3,2 +⋯+t_3,n-1
 = 2n+T_3,n-1
 = 2n+3/4(3ⁿ−2n−1)
 =  n/2 + 3/4(3ⁿ − 1)

 2a = 1 + p and g² = P
⇒ g² = 2a − 1
⇒ 1 − 2a + g² = 0