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

$$1.3+2.3^2+3.3^3+...+n.3^n$$

$$=\dfrac{(2n-1)3^{n+1}+3}{4}$$

Asked by Abhisek | 1 year ago |  65

##### Solution :-

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

P(n): $$1.3+2.3^2+3.3^3+...+n.3^n$$

$$=\dfrac{(2n-1)3^{n+1}+3}{4}$$

For n=1,

L.H.S = 1.31=3

R.H.S $$=\dfrac{(2.1-1)3^{1+1}+3}{4}$$

$$=\dfrac{3^2+3}{4}$$

$$\dfrac{12}{4}=3$$

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

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

$$1.3+2.3^2+3.3^3+...+k.3^k$$

$$=\dfrac{(2k-1)3^{k+1}+3}{4}$$ .........(i)

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

Consider

$$1.3+2.3^2+3.3^3+...+k.3^k+(k+1).3^{k+1}$$

$$1.3+2.3^2+3.3^3+...+k.3^k+(k+1).3^{k+1}$$

$$\dfrac{(2k-1)3^{k+1}+3}{4}+(k+1)3^{k-1}$$......... [Using (i)]

$$\dfrac{(2k-1)3^{k+1}+3+4(k+1)3^{k+1}}{4}$$

$$\dfrac{3^{k+1}(6k+3)+3}{4}$$

$$\dfrac{3^{k+1}.3(2k+1)+3}{4}$$

$$\dfrac{(3^{k+1)+1}(2k+1)+3}{4}$$

$$\dfrac{\{2({k+1})-1\}3^{(k+1)+1}+3}{4}$$

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 Abhisek | 1 year ago

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