मराठी

Show that n55+n33+7n15 is a natural number for all n ∈ N.

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प्रश्न

Show that `n^5/5 + n^3/3 + (7n)/15` is a natural number for all n ∈ N.

सिद्धांत
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उत्तर

Let P(n): `n^5/5 + n^3/3 + (7n)/15`, ∀ n ∈ N.

Step 1: P(1): `1^5/5 + 1^3/3 + (71)/15`

= `(3 + 5 + 7)/15`

= `15/13`

= 1 ∈ N

Which is true for P(1).

Step 2: P(k): `k^5/5 + k^3/3 + (7.k)/15`

Let it be true for P(k) and let `k^5/5 + k^3/3 + (7k)/15` = λ.

Step 3: P(k + 1) = `(k + 1)^5/5 + (k + 1)^3/3 + (7(k + 1))/15`

= `1/5 [k^5 + 5k^4 + 10k^3 + 10k^2 + 5k + 1] + 1/3 [k^3 + 3k^2 + 3k + 1]`

= `(k^5/5 + k^3/3 + (7k)/15) + (k^4 + 2k^3 + 2k) + 1/5 + 1/3 + 7/15 + 71/15  k + 7/15`

= `lambda + k^4 + 2k^3 + 3k^2 + 2k + 1`  ....[From step 2]

= Positive integers

= Natural number

Which is true for P(k + 1).

Hence, P(k + 1) is true whenever P(k) is true.

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पाठ 4: Principle of Mathematical Induction - Exercise [पृष्ठ ७२]

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एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 11
पाठ 4 Principle of Mathematical Induction
Exercise | Q 23 | पृष्ठ ७२

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