Question

If the set of natural numbers is partitioned into subsets ${S_1} = \{ 1\} ,{S_2} = \{ 2,3\} ,{S_3} = \{ 4,5,6\} .........$ and so on. Then the sum of the terms in ${S_{50}}$ is
A. $62525$
B. $25625$
C. $62500$
D. None of these

Answer 1

Hint: Write the last term of each set in the form of the sum of natural numbers then apply the formula of sum of natural numbers to find the last term of ${50^{th}}$ set by putting $n = 50$. In last to fund the sum of ${50^{th}}$ set apply A.P. formula.

Formula Used:
Sum of $n$ natural numbers –
${S_n} = \dfrac{{n(n + 1)}}{2}$
Sum of Arithmetic progress (A.P.) –
$S = \dfrac{n}{2}(2a + (n - 1)d)$
Here, $a$ is the first term of the sequence. Also, the common difference between the terms is $d$.

Complete step by step solution:
Given that,
${S_1} = \{ 1\} ,{S_2} = \{ 2,3\} ,{S_3} = \{ 4,5,6\} .........$
Last term of ${S_1} = 1$
Last term of ${S_2} = 1 + 2$
Last term of ${S_3} = 1 + 2 + 3$
.
.
.
Therefore, the last term of ${n^{th}}$set will be
${S_n} = 1 + 2 + 3 + ........... + n$
${S_n} = \dfrac{{n(n + 1)}}{2} - - - - - (1)$(Sum of $n$ natural numbers)
Last term of ${50^{th}}$set$ = \dfrac{{50(50 + 1)}}{2} = 1275$
Last term of ${49^{th}}$ set$ = \dfrac{{49(49 + 1)}}{2} = 1225$
So, the first term of ${50^{th}}$set$ = 1225 + 1 = 1226$
$ \Rightarrow {S_{50}} = \{ 1225,1226,1227,.........,1275\} $ which is in A.P.
Here, $a = 1225,d = 1226,n = 50$
Using Sum of A.P. formula,
$S = \dfrac{n}{2}(2a + (n - 1)d)$
$S = \dfrac{{50}}{2}(2(1226) + (50 - 1)1)$
$S = 25(2452 + 49)$
$S = 25(2501)$
$S = 62525$
Hence, The Sum of the terms in ${S_{50}}$ is $62525$

Option ‘A’ is correct

Note: In the ${50^{th}}$ try to write at least three numbers in the set. To know the common difference, check the sequence carefully. Observe the number of elements of the consecutive groups which form an A.P.