हिंदी

Prove by induction that for all natural number n sinα + sin(α + β) + sin(α + 2β)+ ... + sin(α + (n – 1)β) = sin(α+n-12β)sin(nβ2)sin(β2)

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प्रश्न

Prove by induction that for all natural number n sinα + sin(α + β) + sin(α + 2β)+ ... + sin(α + (n – 1)β) = `(sin (alpha + (n - 1)/2 beta)sin((nbeta)/2))/(sin(beta/2))`

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उत्तर

Consider P(n): sinα + sin(α + β) + sin(α + 2β) + ... + sin(α + (n – 1)β)

= `(sin (alpha + (n - 1)/2 beta)sin((nbeta)/2))/(sin(beta/2))`, for all natural number n.

We observe that P(1) is true.

Since P(1): sin α = `(sin(alpha + 0) sin beta/2)/(sin  beta/2)`

Assume that P(n) is true for some natural numbers k.

i.e., P(k): sin α + sin(α + β) + sin(α + 2β) + ... + sin(α + (k – 1)β)

= `(sin (alpha + (k - 1)/2 beta)sin((kbeta)/2))/(sin(beta/2))`

Now, to prove that P(k + 1) is true.

We have P(k + 1): sin α + sin(α + β) + sin(α + 2β) + ... + sin(α + (k – 1)β) + sin(α + kβ)

= `(sin (alpha + ("k" - 1)/2 beta)sin((kbeta)/2))/(sin(beta/2)) + sin(alpha + kbeta)`

= `(sin(alpha + (k - 1)/2 beta) sin  (kbeta)/2 + sin(alpha + kbeta) sin  beta/2)/(sin  beta/2)`

= `(cos(alpha - beta/2) - cos(alpha + kbeta - beta/2) + cos(alpha + kbeta - beta/2) - cos(alpha + kbeta + beta/2))/(2sin  beta/2)`

= `(cos(alpha - beta/2) - cos(alpha + kbeta + beta/2))/(2sin  beta/2)`

= `(sin  (alpha + (kbeta)/2)sin  ((kbeta + beta)/2))/(sin  beta/2)`

= `(sin(alpha + (kbeta)/2) sin(k + 1)(beta/2))/(sin  beta/2)`

Thus P(k + 1) is true whenever P(k) is true.

Hence, by the Principle of Mathematical Induction P(n) is true for all natural number n.

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अध्याय 4: Principle of Mathematical Induction - Solved Examples [पृष्ठ ६५]

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एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 11
अध्याय 4 Principle of Mathematical Induction
Solved Examples | Q 8 | पृष्ठ ६५

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