求系列1+11+111+1111+的和…。。最多n个术语

在这里,我们将找到系列1+11+111+1111+的和…。。最多N个术语(其中N是给定的)。 例子:

null
Input : 3Output : 1 + 11 + 111 +....Total sum is : 123Input : 4Output : 1 + 11 + 111 + 1111 +..... Total sum is : 1234Input : 7Output : 1 + 11 + 111 + 1111 + 11111 +          111111 + 1111111 +..... Total sum is : 1234567

这里我们看到,当N的值为3时,级数持续到1+11+111,即三项,其和为123。 求上述级数之和的程序:

C++

// C++ program to find the sum of
// the series 1+11+111+1111+....
#include <bits/stdc++.h>
using namespace std;
// Function for finding summation
int summation( int n)
{
int sum = 0, j = 1;
for ( int i = 1; i <= n; i++) {
sum = sum + j;
// Appending a 1 at the end
j = (j * 10) + 1;
}
return sum;
}
// Driver Code
int main()
{
int n = 5;
cout << " " <<  summation(n);
return 0;
}
// This code is contributed by shivanisinghss2110


C

// C program to find the sum of
// the series 1+11+111+1111+....
#include <stdio.h>
// Function for finding summation
int summation( int n)
{
int sum = 0, j = 1;
for ( int i = 1; i <= n; i++) {
sum = sum + j;
// Appending a 1 at the end
j = (j * 10) + 1;
}
return sum;
}
// Driver Code
int main()
{
int n = 5;
printf ( "%d" , summation(n));
return 0;
}


JAVA

// Java program to find the sum of
// the series 1+11+111+1111+....
import java.io.*;
class GFG
{
// Function for finding summation
static int summation( int n)
{
int sum = 0 , j = 1 ;
for ( int i = 1 ; i <= n; i++)
{
sum = sum + j;
j = (j * 10 ) + 1 ;
}
return sum;
}
// Driver Code
public static void main(String args[])
{
int n = 5 ;
System.out.println(summation(n));
}
}
// This code is contributed
// by Nikita Tiwari


python

# Python program to get the summation
# of following series
def summation(n):
sum = 0
j = 1
for i in range ( 1 , n + 1 ):
sum = sum + j
j = (j * 10 ) + 1
return sum
# Driver Code
n = 5
print (summation(n))


C#

// C# program to find the sum of
// the series 1+11+111+1111+....
using System;
class GFG
{
// Function for finding summation
static int summation( int n)
{
int sum = 0, j = 1;
for ( int i = 1; i <= n; i++)
{
sum = sum + j;
j = (j * 10) + 1;
}
return sum;
}
// Driver Code
public static void Main()
{
int n = 5;
Console.WriteLine(summation(n));
}
}
// This code is contributed by vt_m


PHP

<?php
// PHP program to find the sum of
// the series 1+11+111+1111+....
// Function for finding summation
function summation( $n )
{
$sum = 0; $j = 1;
for ( $i = 1; $i <= $n ; $i ++)
{
$sum = $sum + $j ;
// Appending a 1 at the end
$j = ( $j * 10) + 1;
}
return $sum ;
}
// Driver Code
$n = 5;
echo summation( $n );
// This code is contributed by ajit
?>


Javascript

<script>
// Javascript program to find the sum of
// the series 1+11+111+1111+....
// Function for finding summation
function summation( n) {
let sum = 0, j = 1;
for ( let i = 1; i <= n; i++) {
sum = sum + j;
// Appending a 1 at the end
j = (j * 10) + 1;
}
return sum;
}
// Driver Code
let n = 5;
document.write(summation(n));
// This code contributed by Princi Singh
</script>


输出:

12345

另一种方法: 设一系列S=1+11+111+1111+…+直到第n学期。用公式求级数之和。

\ S = 1 + 11 + 111 + 1111 + . . . + upto  n  term \ \ S = left ( frac{1}{9} 
ight )*left ( 9 + 99 + 999 + 9999 + . . . + upto  n  term 
ight ) \ \ S = left ( frac{1}{9} 
ight ) * left ( left (10 ^{1} - 1
ight ) + left (10 ^{2} -1 
ight ) +left (10 ^{3} -1 
ight ) + left (10 ^{4} -1 
ight ) + . . . + left (10 ^{n} -1 
ight ) 
ight ) \ \ S = left ( frac{1}{9} 
ight ) *left (10 ^{1} + 10 ^{2} + 10 ^{3} + 10 ^{4} + . . . + 10 ^{n} - n
ight ) \ \ S = left ( frac{1}{9} 
ight ) * left ( frac{10 * (10^{n} - 1)}{10 - 1} - n 
ight ) \ \ S = frac{10^{n+1} - 10 - 9n}{81}

下面是上述方法的实现。

C++

// C++ program to find the sum of
// the series 1+11+111+1111+....
#include <bits/stdc++.h>
// Function for finding summation
int summation( int n)
{
int sum;
sum = ( pow (10, n + 1) -
10 - (9 * n)) / 81;
return sum;
}
// Driver Code
int main()
{
int n = 5;
printf ( "%d" , summation(n));
return 0;
}


JAVA

// java program to find the sum of
// the series 1+11+111+1111+....
import java.io.*;
class GFG {
// Function for finding summation
static int summation( int n)
{
int sum;
sum = ( int )(Math.pow( 10 , n + 1 ) -
10 - ( 9 * n)) / 81 ;
return sum;
}
// Driver Code
public static void main (String[] args)
{
int n = 5 ;
System.out.println(summation(n));
}
}
// This code is contributed by anuj_67.


Python3

# Python3 program to
# find the sum of
# the series 1+11+111+1111+....
import math
# Function for
# finding summation
def summation(n):
return int (( pow ( 10 , n + 1 ) -
10 - ( 9 * n)) / 81 );
# Driver Code
print (summation( 5 ));
# This code is contributed
# by mits.


C#

// C# program to find the sum of
// the series 1+11+111+1111+....
using System;
class GFG {
// Function for finding summation
static int summation( int n)
{
int sum;
sum = ( int )(Math.Pow(10, n + 1) -
10 - (9 * n)) / 81;
return sum;
}
// Driver Code
public static void Main ()
{
int n = 5;
Console.WriteLine(summation(n));
}
}
// This code is contributed by anuj_67.


PHP

<?php
//PHP program to find the sum of
// the series 1+11+111+1111+....
// Function for finding summation
function summation( $n )
{
$sum ;
$sum = (pow(10, $n + 1) -
10 - (9 * $n )) / 81;
return $sum ;
}
// Driver Code
$n = 5;
echo summation( $n );
// This code is contributed by aj_36
?>


Javascript

<script>
// javascript program to find the sum of
// the series 1+11+111+1111+....
// Function for finding summation
function summation( n)
{
let sum;
sum = (Math.pow(10, n + 1) -
10 - (9 * n)) / 81;
return sum;
}
// Driver Code
let n = 5;
document.write(summation(n)) ;
// This code is contributed by aashish1995
</script>


输出:

12345

© 版权声明
THE END
喜欢就支持一下吧
点赞7 分享