Given two arrays A and B of same size m. You have to find the sum of nth terms of Fibonacci like series (value of every term is sum of previous two terms) formed by every element of A as first and every element of B as second.

**Examples:**

Input : {1, 2, 3}, {4, 5, 6}, n = 3 Output : 63 Explanation : A[] = {1, 2, 3}; B[] = {4, 5, 6}; n = 3; All the possible series upto 3rd terms are: We have considered every possible pair of A and B and generated third term using sum of previous two terms. 1, 4, 5 1, 5, 6 1, 6, 7 2, 4, 6 2, 5, 7 2, 6, 8 3, 4, 7 3, 5, 8 3, 6, 9 sum = 5+6+7+6+7+8+7+8+9 = 63 Input : {5, 8, 10}, {6, 89, 5} Output : 369

The** naive approach** is to take every pair of the array A an B and make a Fibonacci series with them.

An **efficient approach ** is based on below idea.

Store the original Fibonacci series in an array and multiply the first term by original_fib[n-2] and second term by original_fib[n-1].

Every element of array A, as well as B, will come m times so multiply them by m.

(m * (B[i] * original_fib[n-1]) ) + (m * (A[i] * original_fib[n-2]) )

By using the efficient method it can be written as

original_fib[]={0, 1, 1, 2, 3, 5, 8}; A[] = {1, 2, 3}; B[] = {4, 5, 6}; n = 3; for (i to m) sum = sum + 3*(B[i]*original_fib[2]) + 3*(A[i]*original_fib[1])

Below is the implementation of above approach:

## C++

`// CPP program to find sum of n-th terms ` `// of a Fibonacci like series formed using ` `// first two terms of two arrays. ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `int` `sumNth(` `int` `A[], ` `int` `B[], ` `int` `m, ` `int` `n) ` `{ ` ` ` ` ` `int` `res = 0; ` ` ` ` ` `// if sum of first term is required ` ` ` `if` `(n == 1) { ` ` ` `for` `(` `int` `i = 0; i < m; i++) ` ` ` `res = res + A[i]; ` ` ` `} ` ` ` ` ` `// if sum of second term is required ` ` ` `else` `if` `(n == 2) { ` ` ` `for` `(` `int` `i = 0; i < m; i++) ` ` ` `res = res + B[i] * m; ` ` ` `} ` ` ` ` ` `else` `{ ` ` ` `// fibonacci series used to find the ` ` ` `// nth term of every series ` ` ` `int` `f[n]; ` ` ` `f[0] = 0, f[1] = 1; ` ` ` `for` `(` `int` `i = 2; i < n; i++) ` ` ` `f[i] = f[i - 1] + f[i - 2]; ` ` ` ` ` `for` `(` `int` `i = 0; i < m; i++) { ` ` ` ` ` `// as every b[i] term appears m times and ` ` ` `// every a[i] term also appears m times ` ` ` `res = res + (m * (B[i] * f[n - 1])) + ` ` ` `(m * (A[i] * f[n - 2])); ` ` ` `} ` ` ` `} ` ` ` ` ` `return` `res; ` `} ` ` ` `int` `main() ` `{ ` ` ` `// m is the size of the array ` ` ` `int` `A[] = { 1, 2, 3 }; ` ` ` `int` `B[] = { 4, 5, 6 }; ` ` ` `int` `n = 3; ` ` ` `int` `m = ` `sizeof` `(A)/` `sizeof` `(A[0]); ` ` ` `cout << sumNth(A, B, m, n); ` ` ` `return` `0; ` `} ` |

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## Java

`// Java program to find sum of n-th terms ` `// of a Fibonacci like series formed using ` `// first two terms of two arrays. ` ` ` `public` `class` `GFG { ` ` ` ` ` `static` `int` `sumNth(` `int` `A[], ` `int` `B[], ` `int` `m, ` `int` `n) ` ` ` `{ ` ` ` ` ` `int` `res = ` `0` `; ` ` ` ` ` `// if sum of first term is required ` ` ` `if` `(n == ` `1` `) { ` ` ` `for` `(` `int` `i = ` `0` `; i < m; i++) ` ` ` `res = res + A[i]; ` ` ` `} ` ` ` ` ` `// if sum of second term is required ` ` ` `else` `if` `(n == ` `2` `) { ` ` ` `for` `(` `int` `i = ` `0` `; i < m; i++) ` ` ` `res = res + B[i] * m; ` ` ` `} ` ` ` ` ` `else` `{ ` ` ` `// fibonacci series used to find the ` ` ` `// nth term of every series ` ` ` `int` `f[] = ` `new` `int` `[n]; ` ` ` `f[` `0` `] = ` `0` `; ` ` ` `f[` `1` `] = ` `1` `; ` ` ` `for` `(` `int` `i = ` `2` `; i < n; i++) ` ` ` `f[i] = f[i - ` `1` `] + f[i - ` `2` `]; ` ` ` ` ` `for` `(` `int` `i = ` `0` `; i < m; i++) { ` ` ` ` ` `// as every b[i] term appears m times and ` ` ` `// every a[i] term also appears m times ` ` ` `res = res + (m * (B[i] * f[n - ` `1` `])) + ` ` ` `(m * (A[i] * f[n - ` `2` `])); ` ` ` `} ` ` ` `} ` ` ` ` ` `return` `res; ` ` ` `} ` ` ` ` ` ` ` `public` `static` `void` `main(String args[]) ` ` ` `{ ` ` ` `// m is the size of the array ` ` ` `int` `A[] = { ` `1` `, ` `2` `, ` `3` `}; ` ` ` `int` `B[] = { ` `4` `, ` `5` `, ` `6` `}; ` ` ` `int` `n = ` `3` `; ` ` ` `int` `m = A.length; ` ` ` `System.out.println(sumNth(A, B, m, n)); ` ` ` ` ` `} ` ` ` `// This code is contributed by ANKITRAI1 ` `} ` |

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## Python3

`# Python3 program to find sum of ` `# n-th terms of a Fibonacci like ` `# series formed using first two ` `# terms of two arrays. ` `def` `sumNth(A, B, m, n): ` ` ` ` ` `res ` `=` `0` `; ` ` ` ` ` `# if sum of first term is required ` ` ` `if` `(n ` `=` `=` `1` `): ` ` ` `for` `i ` `in` `range` `(m): ` ` ` `res ` `=` `res ` `+` `A[i]; ` ` ` ` ` `# if sum of second term is required ` ` ` `elif` `(n ` `=` `=` `2` `): ` ` ` `for` `i ` `in` `range` `(m): ` ` ` `res ` `=` `res ` `+` `B[i] ` `*` `m; ` ` ` ` ` `else` `: ` ` ` ` ` `# fibonacci series used to find ` ` ` `# the nth term of every series ` ` ` `f ` `=` `[` `0` `] ` `*` `n; ` ` ` `f[` `0` `] ` `=` `0` `; ` ` ` `f[` `1` `] ` `=` `1` `; ` ` ` `for` `i ` `in` `range` `(` `2` `, n): ` ` ` `f[i] ` `=` `f[i ` `-` `1` `] ` `+` `f[i ` `-` `2` `]; ` ` ` ` ` `for` `i ` `in` `range` `(m): ` ` ` ` ` `# as every b[i] term appears m ` ` ` `# times and every a[i] term also ` ` ` `# appears m times ` ` ` `res ` `=` `(res ` `+` `(m ` `*` `(B[i] ` `*` `f[n ` `-` `1` `])) ` `+` ` ` `(m ` `*` `(A[i] ` `*` `f[n ` `-` `2` `]))); ` ` ` ` ` `return` `res; ` ` ` `# Driver code ` ` ` `# m is the size of the array ` `A ` `=` `[` `1` `, ` `2` `, ` `3` `]; ` `B ` `=` `[` `4` `, ` `5` `, ` `6` `]; ` `n ` `=` `3` `; ` `m ` `=` `len` `(A); ` `print` `(sumNth(A, B, m, n)); ` ` ` `# This code is contributed by mits ` |

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## C#

`// C# program to find sum of ` `// n-th terms of a Fibonacci ` `// like series formed using ` `// first two terms of two arrays. ` `using` `System; ` ` ` `class` `GFG ` `{ ` `static` `int` `sumNth(` `int` `[] A, ` `int` `[] B, ` ` ` `int` `m, ` `int` `n) ` `{ ` ` ` ` ` `int` `res = 0; ` ` ` ` ` `// if sum of first term is required ` ` ` `if` `(n == 1) ` ` ` `{ ` ` ` `for` `(` `int` `i = 0; i < m; i++) ` ` ` `res = res + A[i]; ` ` ` `} ` ` ` ` ` `// if sum of second term is required ` ` ` `else` `if` `(n == 2) ` ` ` `{ ` ` ` `for` `(` `int` `i = 0; i < m; i++) ` ` ` `res = res + B[i] * m; ` ` ` `} ` ` ` ` ` `else` ` ` `{ ` ` ` `// fibonacci series used to find ` ` ` `// the nth term of every series ` ` ` `int` `[] f = ` `new` `int` `[n]; ` ` ` `f[0] = 0; ` ` ` `f[1] = 1; ` ` ` `for` `(` `int` `i = 2; i < n; i++) ` ` ` `f[i] = f[i - 1] + f[i - 2]; ` ` ` ` ` `for` `(` `int` `i = 0; i < m; i++) ` ` ` `{ ` ` ` ` ` `// as every b[i] term appears m ` ` ` `// times and every a[i] term also ` ` ` `// appears m times ` ` ` `res = res + (m * (B[i] * f[n - 1])) + ` ` ` `(m * (A[i] * f[n - 2])); ` ` ` `} ` ` ` `} ` ` ` ` ` `return` `res; ` `} ` ` ` `// Driver Code ` `public` `static` `void` `Main(String[] args) ` `{ ` ` ` `// m is the size of the array ` ` ` `int` `[] A = { 1, 2, 3 }; ` ` ` `int` `[] B = { 4, 5, 6 }; ` ` ` `int` `n = 3; ` ` ` `int` `m = A.Length; ` ` ` `Console.WriteLine(sumNth(A, B, m, n)); ` `} ` `} ` ` ` `// This code is contributed ` `// by Kirti_Mangal ` |

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## PHP

`<?php ` `// PHP program to find sum of n-th terms ` `// of a Fibonacci like series formed using ` `// first two terms of two arrays. ` `function` `sumNth(&` `$A` `, &` `$B` `, &` `$m` `, &` `$n` `) ` `{ ` ` ` ` ` `$res` `= 0; ` ` ` ` ` `// if sum of first term is required ` ` ` `if` `(` `$n` `== 1) ` ` ` `{ ` ` ` `for` `(` `$i` `= 0; ` `$i` `< ` `$m` `; ` `$i` `++) ` ` ` `$res` `= ` `$res` `+ ` `$A` `[` `$i` `]; ` ` ` `} ` ` ` ` ` `// if sum of second term is required ` ` ` `else` `if` `(` `$n` `== 2) ` ` ` `{ ` ` ` `for` `(` `$i` `= 0; ` `$i` `< ` `$m` `; ` `$i` `++) ` ` ` `$res` `= ` `$res` `+ ` `$B` `[` `$i` `] * ` `$m` `; ` ` ` `} ` ` ` ` ` `else` ` ` `{ ` ` ` `// fibonacci series used to find ` ` ` `// the nth term of every series ` ` ` `$f` `= ` `array` `(); ` ` ` `$f` `[0] = 0; ` ` ` `$f` `[1] = 1; ` ` ` `for` `(` `$i` `= 2; ` `$i` `< ` `$n` `; ` `$i` `++) ` ` ` `$f` `[` `$i` `] = ` `$f` `[` `$i` `- 1] + ` `$f` `[` `$i` `- 2]; ` ` ` ` ` `for` `(` `$i` `= 0; ` `$i` `< ` `$m` `; ` `$i` `++) ` ` ` `{ ` ` ` ` ` `// as every b[i] term appears m times ` ` ` `// and every a[i] term also appears m times ` ` ` `$res` `= ` `$res` `+ (` `$m` `* (` `$B` `[` `$i` `] * ` `$f` `[` `$n` `- 1])) + ` ` ` `(` `$m` `* (` `$A` `[` `$i` `] * ` `$f` `[` `$n` `- 2])); ` ` ` `} ` ` ` `} ` ` ` ` ` `return` `$res` `; ` `} ` ` ` `// Driver code ` ` ` `// m is the size of the array ` `$A` `= ` `array` `(1, 2, 3 ); ` `$B` `= ` `array` `(4, 5, 6 ); ` `$n` `= 3; ` `$m` `= sizeof(` `$A` `); ` `echo` `(sumNth(` `$A` `, ` `$B` `, ` `$m` `, ` `$n` `)); ` ` ` `// This code is contributed ` `// by Shivi_Aggarwal ` `?> ` |

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**Output:**

63