Home
Class 12
MATHS
The sequence of odd natural numbers is d...

The sequence of odd natural numbers is divided into groups `1,3,5,7,9,11,"…."` and so on. Show that the sum of the numbers in nth group is `n^(3)`.

Text Solution

Verified by Experts

Sequence of natural number is divided into group 1,3,5,7,9,11…
` therefore ` nth row contains n elements
1st elements of nth row =`n^(2) -( n-1) `
Least elements of nth row ` =n^(2) +( n-1) `
` therefore `Sum of the element in the nth row ltbr gt ` = (n)/(2) (a+l) =(n)/(2) [n^(2) + n-1] =(n)/(2) [2n ^(2) ] ]=n^(3) `
` =(n)/(2) [n^(2) -n+1+ n^(2) + n-1])(n)/(2) [2n^(2) ] =n^(3)`.
Promotional Banner

Similar Questions

Explore conceptually related problems

If the natural numbers are divided into the groups (1),(2,3),(4,5,6),(7,8,9,10) .. then the sum of the elements of 25^(th ) group is

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 find the sum of the terms in S_(50)dot

Find the sum of first n odd natural numbers.

The natural numbers arearranged innthe form given below The rth group containing 2^(r-1) numbers. Prove that sum of the numbers in the nth group is 2^(n-2)[2^(n)+2^(n-1)-1] .

Find the sum of natural odd numbers from 1 to 100.

If N, the set of natural numbers is partitioned into groups S_(1)={1},S_(2)={2,3},S_(3)={4,5,6,7},S_(4)={8,9,10,11,12,13,14,15},"....", find the sum of the numbers in S_(50).

If the sum of first n odd natural number is 1225, find the value of n.

Find the sum of all two digit natural numbers which when divided by 3 yield 1 as remainder.

Let n be an odd natural number of greater than 1. then the number of zeros at the end of the sum 999^(n)+1 is:

A six faced dice is so biased that it is twice as likely to show an even number as an odd number when thrown . It is thrown twice . The probability that the sum of two numbers thrown is even is :