. Ex 1) Find t and the sum of n terms of
1, 4, 12, 32, 80, 192, ...
Please answer this step by step and send me attachments
Answers
Given : 1, 4, 12, 32, 80, 192, ...
To find : nth term (Tₙ) and sum of n terms ∑Tₙ ( 1 to n)
Solution:
1, 4, 12, 32, 80, 192, ...
1 = 2⁰ × 1
4 = 2¹ × 2
12 = 2² × 3
32 = 2³× 4
80 = 2⁴× 5
192 = 2⁵× 6
Next term = 2⁶ x 7 = 448
Tₙ = 2ⁿ⁻¹ * n
∑ n(2ⁿ⁻¹)
2⁰ × 1 + 2¹ × 2 + 2² × 3 + .........................+ 2ⁿ⁻¹ * n
= (2⁰ + 2¹ + .................+2ⁿ⁻¹ ) + (2¹ + 2²+ ... .+2ⁿ⁻¹ ) + ...................... + ( 2ⁿ⁻¹)
Sₙ = a(rⁿ-1)/ (r - 1)
2⁰ + 2¹ + .................+2ⁿ⁻¹ = 2⁰(2ⁿ-1) /(2 -1) = 2ⁿ - 2⁰
(2¹ + 2²+ ... .+2ⁿ⁻¹ ) = 2¹(2ⁿ⁻¹ - 1) /(2 -1) = 2ⁿ⁻¹ = 2ⁿ - 2¹
2ⁿ - 2²
and so on
nth term 2ⁿ⁻¹ = 2ⁿ -2ⁿ⁻¹
∑ n(2ⁿ⁻¹)
= 2ⁿ - 2⁰ + 2ⁿ - 2¹ + 2ⁿ - 2² +.............................................+ 2ⁿ -2ⁿ⁻¹
= n2ⁿ - ( 2⁰ + 2¹ + ..........................+ 2ⁿ⁻¹ )
= n2ⁿ - ( 2ⁿ - 1)
= n2ⁿ - 2ⁿ + 1
= 2ⁿ (n - 1) + 1
Tₙ = 2ⁿ⁻¹ * n
sum of n terms of 1, 4, 12, 32, 80, 192, ... = ∑ n(2ⁿ⁻¹) = n2ⁿ - 2ⁿ + 1
= 2ⁿ (n - 1) + 1
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