<p>1/(12) , 1/(23) , 1/(34)
The first three terms of a sequence are given above. The nth term of the sequence is 1/(n(n+1)), which is equal to (1/n)-(1/(n+1)). What is the sum of the first 50 terms of this sequence?
A. 1
B. 50/51
C. 49/50
D. 24/50
E. 1/(5051)</p>
<p>The answer however is B. Why?</p>
<p>As you may see, the first term is 1- 1/2, the 2nd term is 1/2 - 1/3 and for forth</p>
<p>So the 50th term is 1/50 - 1/51</p>
<p>If you add up the first 50 term you get 1 - 1/2 + (1/2 - 1/3) … + 1/50 - 1/51
all the numbers between 1/2 to 1/50 will be counterbalanced.
so the result is 1- 1/51 and you got 50/51</p>
<p>Let me know if you have any questions.</p>
<p>i solved it in this way but i don’t know whether is way of solving is accurate or not </p>
<p>1 / n(n+1 ) = 1 / n - 1/n+1</p>
<p>n(n+1)/ n - n(n+1) / n +1 = 0</p>
<p>n+1 = n </p>
<p>n / n +1 or n+1 /n</p>
<p>So they will all counterbalance? Why is that? Is that supposed to be a general rule?</p>
<p>That’s not a rule, if you add up negative 1/2 and positive 1/2, you get zero; and f you add up negative 1/3 and positive 1/3 in the equation you get zero as well. Try to list the sum of the 50 term, and watch out to the negative and positive numbers, you should be able to deduce that there are only two numbers left, 1 and negative 1/51.</p>