Question

The 4800th term of the sequence,1,3,3,3,5,5,5,5,5,7,7,7,7,7,7,7,...is

Answer 1

Answer:

139.....................

Answer 2

Concept

The series special in some way or the other is called a special series. The following are the three types of special series.

• 1 + 2 + 3 +… + n (sum of first n natural numbers)

• $1 ^2+ 2^2 + 3^2 +... + n^2$ (sum of squares of the first n natural numbers)

• $1^3 + 2^3 + 3^3 +...+ n^3$ (sum of cubes of the first n natural numbers)

Given

We have given a series 1,3,3,3,5,5,5,5,7,7,7,7,7,7,7,7,7,7,7,7,7

Find

We are asked to determine the 4800th term of the sequence

Solution

Label the terms of the sequence $a_1,a_2,a_3,......a_4_8_0_0 .....$

Notice that the last term with the value 1 is $a_1$, the last term with the value 3 is $a_3$ , the last term with the value 5 is $a_9$, and in general, the last term with the value $2k-1$ is $a_{s_{k} }$ where

$S_k=\sum\limit_{i=1} ^k(2i-1)=1+3+5+7+.....+(2k-1)$ .

Then, to know the value of the terms in the subsequence containing $a_{4800}$, we need only find k such that $S_{k-1} < 4800S_k$ Then we will have $a_{4800} =2k-1$. To do so, we will put $S_k$ into closed form.

$S_k=\sum\limit_{i=1} ^k(2i-1)\\\\=-k+2\sum\limit _{i=1}^k(i)\\\\=-k+2\frac{k(k+1)}{2} \\\\=-k+k^2+k\\\\=k^2$

Examining squares, we find that

$S_6_9=69^2 < 4800\leq 70^2=S_7_0$

Thus $a_{4800}$ occurs after the final term of the subsequence consisting of $2(69)-1=137$ but before the last term of the subsequence consisting of $2(70)-1=139$ giving us $139$

Therefore, the 4800th term is 139 .

#SPJ2