If sequence T is defined for all positive integers n such that tn +1 =

Question

If sequence T is defined for all positive integers n such that  t_{(n +1)} = t_{n} + n , and  t_3 = 14 , what is  t_{20} ?

A. 101 
B. 187 
C. 191 
D. 201 
E. 251

KarishmaB · Accepted Answer

Please ensure that the question is typed clearly and has no ambiguity. You are given t(n +1) = t(n) + n not t(n +1) = t(n + n)

t_{n+1} = t_{n} + n 
 t_{20} = t_{19} + 19 
 t_{20} = t_{18} + 18 + 19

We see the pattern:
 t_{20} = t_3 + 3 + 4 + 5 + 6 + ... + 17 + 18 + 19

t_{20} = 14 + 3 + 4 + 5 + ... + 18 + 19  
Split 14 into 11 + 1 + 2 for ease of calculation:

t_{20} = 11 + 1 + 2 + 3 + 4 + ... + 18 + 19 
Sum of first m positive integers is m(m+1)/2 so

t_{20} = 11 + 19*20/2 = 201

Answer (D)

hycday · Answer

ok, not sure if this is the best explanation, but here is my way of solving this :

tn +1 = tn + n, 
t 3 = 14
so:
t 4 = t 3 + 3
t 5 = t 4 + 4, so we replace t 4 with the previous result : t 5 = (t 3 + 3 ) + 4
....
so we get
t 20 = t 3 + 3 + 4 + 5 + 6....+ 19
the 3 + 4 + 5 + ... + 19 sequence is equal the sequence 1 + 2 + ... + 19, minus 1 + 2 at the beginning (so, minus 3)
and the 1 + 2 ... + 19 sequence is equal to n*(n+1)/2 with n being 19, so equal to 19 * 20 / 2 = 190
so :
t 20 = t 3 + 190 - 3 
t 20 = 14 + 190 - 3 = 190 + 11 = 201, hence answer D

BrainFog · Answer

Calculate the first few values to see a pattern

t1= 11
t2= 12
t3= 14
t4= 17
t5= 21

The sequence is basically like this t1 + 1 +2 +3 +4 +5 ...

So t20 = t1 + 1 +2 +3 up to 19 = 11 +   * 19 = 201

Hope this helps. A big thanks to  Bunuel , I learned this technique reading his posts.

BillyZ · Answer

Source : Manhattan Advanced Quant Question No. 8

OFFICIAL SOLUTION

First, we can write a few of the terms, noting the relationship of each term to the previous one:
 
 t_3 = 14 
 t_4 = 14 + 3 
 t_5 = 14 + 3 + 4 
 t_6 = 14 + 3 + 4 + 5 
…
 t_{20} = 14 + (3 + 4 + 5 + … + 18 + 19) 
 
To evaluate  t_{20}  , we need to compute the sum contained in parentheses above.
 
We can use the rule that (sum of a set of consecutive integers) = (middle term) × (number of terms).
The middle term can be found by taking the average of the two extreme terms 3 and 19 to get 11. The number of terms is  19 – 3 + 1 = 17 .
Now we can compute  11 × 17 = 187 .
 
Finally, we have  t_{20} = 14 + (3 + 4 + 5 + … + 18 + 19) = 14 + 187 = 201 .
 
The correct answer is D.

Cez005 · Answer

Where does the "11" come from here?

alexiaint · Answer

t_{n+1} = t_{n} + n 
 t_{20} = t_{19} + 19 
 t_{20} = t_{18} + 18 + 19

Also, since  t_{3} = t_{2} + 2 = 14  -> 
 t_{2} = t_{1} + 1  
and  t_{1} =11

The pattern:
 t_{20} = t_1 + (1 +2 + 3 + 4 + 5 + 6 + ... + 17 + 18 + 19)  where t_1 = 11

Sum of numbers from 1 to 19 inclusive is Avg. * # of terms
Avg. = (1 + 19)/2 = 10
Sum = 10*19=190

Thus:
 t_{20} = t_1 + 190 = 11+190 = 201

Answer (D)

matthewsmith_89 · Answer

My approach: t3 = 14 t4 = 14 + 3 = 17 t20 = 14+3 + 4+5+ ... + 19 t20= 11 + 1+ 2 + ... + 19 sum of (1 to 19) = 19/2(19 + 1) = 190 Therefore t20 = 190 + 11 = 201 Which is answer D :-D

Kinshook · Answer

Asked : If sequence T is defined for all positive integers n such that  t_{(n +1)} = t_{n} + n , and  t_3 = 14 , what is  t_{20} ?

t_{n+1} = t_n + n = t_{n-1} + n-1 + n = .... = t_3 + 3 + 4 + 5 + 6 + ... + n = 14 + 1 + 2 + 3 + ... n - 1 - 2 = 11 + \frac{n(n+1)}{2}

t_{20} = 11 + \frac{19*20}{2} = 11 + 190 = 201

IMO D

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If sequence T is defined for all positive integers n such that tn +1 =

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