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How do you find a formula for #sum_(r=0)^n r2^r# ?

Answer 1

Genesis Allen

#sum_(r=0)^n r2^r = (n-1)2^(n+1)+2#

Take note of this:

#(n+1)2^(n+1)+sum_(r=0)^n r2^r = sum_(r=0)^(n+1) r2^r#

#color(white)((n+1)2^(n+1)+sum_(r=0)^n r2^r) = sum_(r=1)^(n+1) r2^r#

#color(white)((n+1)2^(n+1)+sum_(r=0)^n r2^r) = sum_(r=1)^(n+1) 2^r + sum_(r=1)^(n+1) (r-1)2^r#

#color(white)((n+1)2^(n+1)+sum_(r=0)^n r2^r) = (2^(n+2)-2) + 2sum_(r=1)^(n+1) (r-1)2^(r-1)#

#color(white)((n+1)2^(n+1)+sum_(r=0)^n r2^r) = (2^(n+2)-2) + 2sum_(r=0)^n r2^r#

Subtract #sum_(r=0)^n r2^r+(2^(n+2)-2)# from both ends to get:

#sum_(r=0)^n r2^r = (n+1)2^(n+1)-2^(n+2)+2#

#color(white)(sum_(r=0)^n r2^r) = (n-1)2^(n+1)+2#

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Answer 2

Eva Adamson

#2(1 + 2^n (n-1))#

#sum_(r=0)^nrx^r=xsum_(n=0)^nrx^(r-1)=x d/(dx)(sum_(n=0)^nx^r)=# #x d/(dx)((x^(n+1)-1)/(x-1))=(x(1 + (n (x-1)-1) x^(n)))/(x-1)^2#

now making #x=2# we have

#sum_(r=0)^nr2^r=2(1 + 2^n (n-1))#

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Answer 3

Ethan Adkins

The formula for the sum ( \sum_{r=0}^{n} r \cdot 2^r ) is ( n \cdot 2^{n+1} - (n+1) ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.