The r_("th")rth term of a geometrical series is (2r+1)cdot 2^r(2r+1)⋅2r. The sum of the first nn term of the series is what?

2 Answers
Vinícius Ferraz
Jun 13, 2018

(4n-2)*2^n + 3(4n−2)⋅2n+3

Explanation:

S = sum_{r=0}^n 2r * 2^r + sum_{r=0}^n 2^rS=n∑r=02r⋅2r+n∑r=02r

S= sum_{r=1}^n r * 2^(r+1) + (1 - 2^{n+1})/(1 - 2)S=n∑r=1r⋅2r+1+1−2n+11−2

S= a_{01} ( 1 - 2^n)/ (1- 2) + ... + a_{0n} ( 1 - 2^{n-(n-1)})/ (1- 2) + 2^{n+1} - 1

1*2^2 + 1*2^3 + 1*2^4

+ 1 * 2^3 + 1 * 2^4

+ 1 * 2^4

S = sum_{i = 0}^{n-1} 2^{i+2} (2^(n - i) - 1) + 2^{n+1} - 1

S = 4 sum_{i = 0}^{n-1} (2^n - 2^i) + 2^{n+1} - 1

S = 4*2^n * n - 4 * (2^n - 1) + 2^{n+1} - 1

S = (4n-2)*2^n + 3

Let's verify

S = 1 * 2^0 + 3 * 2^1 + 5 * 2^2 + 7 * 2^3 + cdots

S = 1 + 6 + 20 + 56 + cdots

S(0) = 1 = -2 + 3

S(1) = 7 = 2 * 2 + 3

S(2) = 27 = 6 * 2^2 + 3

And S(3) = 83 = 10 * 2^3 + 3

Ratnaker Mehta
Jun 15, 2018

(4n-2)2^n+2, or, (2n-1)2^(n+1)+2

Explanation:

Let S_n denote the sum of the first n terms of the sequence

{(2r+1)2^r | r in NN}.

Then, S_n=sum_(r=1)^(r=n)(2r+1)2^r,

:.S_n=3*2^1+5*2^2+...+(2n-1)*2^(n-1)+(2n+1)*2^n.

Multiplying by 2, we get,

2S_n=3*2^2+5*2^3+...+(2n-1)*2^n+(2n+1)*2^(n+1).

:.2S_n-S_n=(3-5)2^2+(5-7)2^3+...+{(2n-1)-(2n+1)}2^n+(2n+1)2^(n+1)color(red)(-3*2^1).

:.S_n=color(red)(-2*2^1)-2*2^2-2*2^3-2*2^n+(2n+1)2^(n+1)color(red)(-1*2^1),

=-2[2^1+2^2+2^2+2^3+...+2^n]+(2n+1)color(blue)(2^(n+1))-2,

=-2[{2(2^n-1)}/(2-1)]+(2n+1)color(blue)(2^n*2)-2,

=-4*2^n+4+(4n+2)2^n-2.

=2^n{-4+(4n+2)}+4-2.

rArr S_n=(4n-2)2^n+2, or, S_n=(2n-1)2^(n+1)+2

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