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47 votes
47 votes

The recurrence equation
$ T(1) = 1$
$T(n) = 2T(n-1) + n, n \geq 2$

evaluates to

  1. $2^{n+1} - n - 2$
  2. $2^n - n$
  3. $2^{n+1} - 2n - 2$
  4. $2^n + n $
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7 Answers

Best answer
65 votes
65 votes
$T(n) = 2T(n-1) +n, n \geqslant 2 , T(1) = 1$

$  T(n)    = n + 2(n-1) + 2^2 (n-2) + \dots + 2^{(n-1)}(n -(n-1))$

           $=n( 1 + 2 + \dots + 2^{n-1}) - (1.2 + 2.2^{2} +3.2^{3}+\dots+ (n-1).2^{n-1})$

           $=n(2^n-1) -(n.2^n-2^{n+1}+2)$

           $=2^{n+1}-n -2$

Correct Answer: $A$
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105 votes
105 votes
T(1) = 1

T(2) = 4

T(3) = 11

T(4) = 26

T(5) = 57

T(6) = 120

T(7) = 247

So, $$T(n) = 2^{n+1} - n - 2$$
29 votes
29 votes
Given that T(1) =1 //OMG here itself Option C & D fails now check A & B

T(2)=2T(1)+n=2+2=4 //Option B evaluates to 2 so it is wrong

Hence option A is ans.
12 votes
12 votes

solution .

Answer:

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