1

Find the sum of the n terms of the series:

$2\cdot2^0+3\cdot2^1+4\cdot2^2+\dots$

I don't know how to proceed. Please explain the process and comment on technique to solve questions of similar type.

Source: Barnard and Child Higher Algebra.

Thanks in Advance!

Martin Sleziak
  • 53,687
gaufler
  • 1,871
  • This is an Arithmetico--Geometric Progression. – Aditya Agarwal Sep 09 '15 at 10:50
  • 2
    HINT: $\sum\limits_{k=0}^{n}(k+2)\cdot2^k=\left(\color\red{\sum\limits_{k=0}^{n}k\cdot2^k}\right)+\left(\color\green{\sum\limits_{k=0}^{n}2\cdot2^k}\right)$. – barak manos Sep 09 '15 at 11:39
  • 1
    You may find some of these questions useful: http://math.stackexchange.com/questions/87030/proof-by-induction-sum-limits-i-0n-i-2i-1-n1-2n-1 http://math.stackexchange.com/questions/11464/how-to-compute-the-formula-sum-r-1d-r-cdot-2r http://math.stackexchange.com/questions/90637/what-is-the-limit-of-sum-limits-n-1-inftyn2-3n http://math.stackexchange.com/questions/30732/how-can-i-evaluate-sum-n-0-infty-n1xn – Martin Sleziak Sep 09 '15 at 11:46

3 Answers3

4

The formula for the sum of a geometric sequence gives $$ \sum_{k=0}^n2^kx^{k+2}=\frac{x^2-2^{n+1}x^{n+3}}{1-2x}\tag{1} $$ Differentiating $(1)$ yields $$ \sum_{k=0}^n(k+2)2^kx^{k+1}=\frac{2x-(n+3)2^{n+1}x^{n+2}}{1-2x}+\frac{2x^2-2^{n+2}x^{n+3}}{(1-2x)^2}\tag{2} $$ Plugging in $x=1$ leads to $$ \sum_{k=0}^n(k+2)2^k=(n+1)2^{n+1}\tag{3} $$

robjohn
  • 345,667
  • please explain how did you get $\sum_{k=0}^n2^kx^{k+2}=\frac{x^2-2^{n+1}x^{n+3}}{1-2x}\tag{1}$ this. I am unable to get that RHS equation. – gaufler Sep 09 '15 at 12:51
  • 1
    $$\begin{align}\sum_{k=0}^n2^kx^{k+2} &=x^2\sum_{k=0}^n2^kx^k\ &=x^2\frac{1-2^{n+1}x^{n+1}}{1-2x}\ &=\frac{x^2-2^{n+1}x^{n+3}}{1-2x}\end{align}$$ – robjohn Sep 09 '15 at 14:10
2

Hint: Let $S$ be the sum. Can you simplify $2S-S$ by keeping like powers of $2$ together and using the expression for summing a G.P.?

Macavity
  • 46,381
2

This lookslike a double sum. Try rewriting it the following way. $$2\cdot 2^0+3\cdot 2^1+4\cdot 2^2+...+(n+2)\cdot 2^n =$$ $$=2\cdot 2^0+2\cdot 2^1+2\cdot 2^2+...+2\cdot 2^n$$ $$+1\cdot2^1+1\cdot 2^2+...+1\cdot 2^n$$ $$+1\cdot 2^2+...+1\cdot 2^n$$ $$ \cdots $$

The Terms are now a simple geometric series.

MrYouMath
  • 15,833