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Asked by Ritu Last Modified
Answer As an experienced tutor registered on UrbanPro, I can confidently demonstrate that the expression n55+n33+7n155n5+3n3+157n yields a natural number for all nn belonging to the set of natural numbers.
To prove this, let's consider each term of the expression separately:
n555n5: Since n5n5 is always a natural number (since nn is a natural number), dividing it by 5 can result in a fraction, but since 5 is a factor of n5n5, the division yields an integer.
n333n3: Similar to the previous term, n3n3 is always a natural number, and dividing it by 3 might yield a fraction, but since 3 is a factor of n3n3, the division also yields an integer.
7n15157n: Here, 15 is a multiple of both 5 and 3. So, when we divide 7n7n by 15, it will result in an integer, as both 7n7n and 15 are divisible by 7 and 15 respectively.
Since all three terms of the expression yield integers individually, their sum will also be an integer. And since it's a sum of natural numbers, it will be a natural number.
Therefore, n55+n33+7n155n5+3n3+157n is indeed a natural number for all nn belonging to the set of natural numbers. This demonstrates the validity of the expression for all nn in NN.
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