Question

Prove by the principle of mathematical induction that $1+2+3+...+n<\frac{(2n+1{)}^2}{8}$, for all $n\in N.$

Or

Using the principle of mathematical induction, prove that

$1.3+{2.3}^2+{3.3}^3+...+n{.3}^n=\frac{(2n-1)3^{n+1}+3}{4}.$ for all $n\in N.$

Answer 1

Let P(n) be the statement given by

$P\left(n\right):1+2+3+...+n<\frac{(2n+1{)}^2}{8}$

For $n=1,P\left(1\right):1<\frac{(2+1{)}^2}{8}$

$\because 1<\frac{9}{8}$

$\therefore$ P(1) is true.

Assume that, P(k) be true, for some natural number k.

$1+2+3+...+k<\frac{(2k+1{)}^2}{8}$

Now, we will show that P(k+1) is true whenever P(k) is true.

i.e. $1+2+3+...+k+k+1<\frac{{\left\{2\right(k+1)+1\}}^2}{8}$

Now, P(k) is true.

$\Rightarrow 1+2+3+...+k<\frac{(2k+1{)}^2}{8}$

$\Rightarrow 1+2+3+...+k+(k+1)<\frac{(2k+1{)}^2}{8}+(k+1)$

[adding both sides by (k+1)]

$\Rightarrow 1+2+3+...+k+(k+1)<\frac{1}{8}\left\{\right(2k+1{)}^2+8(k+1)\}$

$\Rightarrow 1+2+3+...+k+(k+1)<\frac{1}{8}(4k^2+4k+1+8k+8)$

$\Rightarrow 1+2+3+...+k+(k+1)<\frac{1}{8}(4K^2+12K+9)$

$\Rightarrow 1+2+3+...+K+(K+1)<\frac{1}{8}(2k+3{)}^2$

$\Rightarrow 1+2+3+...+k+(k+1)<\frac{1}{8}{\left\{2\right(k+1)+1\}}^2$

Thus, P(k+1) is true.

Hence, by principle of mathematical induction, P(n) is true, $\forall n\in N.$

Let P(n) be the statement given by

$P\left(n\right):1.3+{2.3}^2+{3.3}^2+...+n{.3}^n=\frac{(2n-1)3^{n+1}+3}{4}$

For $n=1:P\left(1\right):l.3=\frac{\left\{2\right(1)-1\}{3}^{1+1}+3}{4}$

$\Rightarrow 3=\frac{9+3}{4}\Rightarrow 3=3$

So, P(1) is true.

Assume that P(k) be true, for some natural number k.

$\therefore 1.3+{2.3}^2+{3.3}^3+...+k{.3}^k=\frac{(2k-1)3^{k+1}+3}{4}...\left(i\right)$

Now, we will show that $P(k+1)$ is true, whenever P(k) is true

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

= $\frac{\left\{2\right(k+1)-1\}{3}^{(k+1)+1}+3}{4}$

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

= $\frac{(2k-1)3^{k+1}+3}{4}+(k+1){.3}^{k+1}$ [from Eq. (i)]

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

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

= $\frac{(6k+3){.3}^{k+1}+3}{4}$

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

= $\frac{\left\{2\right(k+1)-1\}{.3}^{(k+1)+1}+3}{4}$

$\therefore P(k+1)$ is true.

Hence, by principle of mathematical induction, P(n) is true, $\forall n\in N.$