Prove the following by using the principle of mathematical induction for all $$n\in N$$

$$a+ar+ar^2+....+ar^{n-1}=\dfrac{a(r^n-1)}{r-1}$$

Asked by Abhisek | 1 year ago |  69

##### Solution :-

Let us denote the given equality by P(n), i.e.,

$$a+ar+ar^2+....+ar^{n-1}=\dfrac{a(r^n-1)}{r-1}$$

For n=1,

L.H.S.= a

R.H.S. = $$\dfrac{a(r^n-1)}{r-1}=a$$

Therefore, P(n) is true for n=1.

Let us assume that P(k) is true for some positive integer k , i.e.,

$$a+ar+ar^2+....+ar^{k-1}=\dfrac{a(r^k-1)}{k-1}$$.........(i)

Now, we have to prove that P(k+1) is also true

Consider

$$\{ a+ar+ar^2+....+ar^{k-1}\}+ar^{(k+1)-1}$$

$$\dfrac{a(r^k-1)}{r-1}+ar^k$$ ...........[Using (i)]

$$\dfrac{a(r^k-1)+ar^k(r-1)}{r-1}$$

$$\dfrac{a(r^k-1)+ar^{k+1}-ar^k}{r-1}$$

$$\dfrac{ar^k-a+ar^{k+1}-ar^k}{r-1}$$

$$\dfrac{ar^{k+1}-a}{r-1}$$

$$\dfrac{a(r^{k+1}-1)}{r-1}$$

Therefore, P(k+1) holds whenever P(k) holds.

Hence, the given equality is true for all natural numbers i.e., N by the principle of mathematical induction.

Answered by Pragya Singh | 1 year ago

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