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

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 +1)/2] * 19 = 201

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

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.

Where does the "11" come from here?

\(t_{n+1} = t_{n} + n\)
\(t_{20} = t_{19} + 19\)
\(t_{20} = t_{18} + 18 + 19\)

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)

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

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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