Principle of Mathematical Induction

6. Let the given statement be P (n) i.e.,

P (n)=1.2+2.3+3.4+ … +2 (n+1)= $\left[\frac{n(n+1)(n+2)}{3}\right]$

For n=1,

P (1)=1.2=2= $\frac{1(1+1)(1+2)}{3}$ = $\frac{2*3}{3}$ =2.

Which is true.

considerP (k) be true for some positive integer k

1.2 + 2.3 + 3.4 + … + k (k + 1) = $\left[\frac{k(k+1)(k+2)}{3}\right]$ - (1)

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

Here, 1.2 + 2.3 + 3.4 + … + k (k + 1) + (k+1) (k+2)

By using (1), we get

= $\frac{k(k+1)(k+2)}{3}+(k+1)(k+2)$

= (k+1) (k+2) $\left[\frac{k}{3}+1\right]$

= $\frac{(k+1)(k+2)(k+3)}{3}$

By further simplification; $\frac{(k+1)(k+1+1)(k+1+2)}{3}$

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

Therefore, by the principle of mathematical induction, statement P (n) is true for all natural no. i.e., n.

5. Let the given statement be P(n) i.e.,

P(n)= 1.3 + 2.3² + 3.3³ + … + n.3ⁿ = $\frac{(2n-1)3^{n+1}+3}{4}$

If n=1, we get

P(1) = 1.3=3= $\frac{(2.1-1)3^{1+1}+3}{4}$ = $\frac{1·3^2+3}{4}$ = $\frac{12}{4}$ =3

which is true.

Consider P(k) be true for some positive integer k

1.3 + 2.3² + 3.3³ + … + k^3k = $\frac{(2k-1)3^{k+1}+3}{4}$ ------------------(1)

Now, let us prove P(k+1) is true.

Here,

1.3 + 2.3² + 3.3³ + … + k^3k + (k + 1)3^{k + 1}

By using eqn. (1)

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

L.C.M

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

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

= $\frac{3^{k+1}\{2k-1+4k+4\}+3}{4}$

= $\frac{3^{k+1}\{6k+3\}+3}{4}$

= $\frac{3^{k+1}·3(2k+1)+3}{4}$ = $\frac{3^{k+1+1}\{2k+1\}+3}{4}$ ?P(k+1) is true whenever P(k) is true.

Therefore, by the principle of mathematical induction statement P(n) is true for all natural numbers i.e., n.

4. Let the given statement be P(n) i.e.,

P(n): 1.2.3 + 2.3.4 + … + n (n + 1)(n + 2) = $\frac{n(n+1)(n+2)(n+3)}{4}$

If n=1, we get

P(1): 1.2.3 = 6 = $\frac{1(1+1)(1+2)(1+3)}{4}$ = $\frac{1.2.3.4}{4}=6$

which is true.

considerP(k) is true for some positive integer k

1.2.3 + 2.3.4 + … + k(k + 1)(k + 2) = $\frac{k(k+1)(k+2)(k+3)}{4}$ -------------------(1)

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

Here,1.2.3 + 2.3.4 + … + k(k + 1)(k + 2) + (k + 1)(k + 2)(k + 3)

By eqn (1), we get,

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

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

= $\frac{(k+1)(k+2)(k+3)(k+4)}{4}$

By further Simplification,

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

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

Hence, from the principle of mathematical induction, the P(n) is true for all natural numbers n.

3. Let the given statement be P(n) i.e.,

P(n): 1+ $\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+?+\frac{1}{(1+2+3+?n)}=\frac{2n}{(n+1)}$

For n=1,

we get,P(1)=1= $\frac{2.1}{1+1}=\frac{2}{2}=1$

which is true.

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

i.e., $1+\frac{1}{1+2}+\dots +\frac{1}{(1+2+3+?k)}=\frac{2k}{(k+1)}$ ------------------ (1)

Which is true.

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

$1+\frac{1}{1+2}+\frac{1}{1+2+3}$ + … $+\frac{1}{(1+2+3+?k)}+\frac{1}{1+2+3+?+k+(k+1)}$

By using eqn (1)

= $\frac{2k}{(k+1)}$ + $\frac{1}{(1+2+3+\dots k+1)}$

?We know that, 1+2+3+ … +n= $\frac{n(n+1)}{2}$

So, we get

= $\frac{2k}{k+1}$ + $\frac{1}{\left(\frac{k+1(k+1+1)}{2}\right)}$

= $\frac{2k}{k+1}$ + $\frac{2}{(k+1)(k+2)}$

= $\frac{2}{k+1}\left\{k+\frac{1}{k+2}\right\}$

= $\frac{2}{\left(k+1\right)}\left\{\frac{k(k+2)+1}{k+2}\right\}$

= $\frac{2}{k+1}\left\{\frac{k^2+2k+1}{k+2}\right\}$

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

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

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

Hence, from the principle of mathematical induction, the P(n) is true for all a natural number n.

2. Let the given statement be P(n) i.e.,

P(n)=1³+2³+3³+ … +n³= ${\left(\frac{n(n+1)}{2}\right)}^2$

For, n=1, P(n)=1³=1= ${\left(\frac{1(1+1)}{2}\right)}^2={\left(\frac{1.2}{2}\right)}^2=1^2=1$

which is true.

Consider P(k) be true for some positive integer k

1³+2³+3³+ … +k³= ${\left(\frac{k(k+1)}{2}\right)}^2$ ---------- (1)

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

Here,  1³+2³+3³+ … +k³+(k+1)³

By using eq (1)

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

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

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

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

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

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

Hence, from the principle of mathematical induction, P(n) is true for all natural numbers n.

1. Let the given statement be P(n) i.e.,

P(n): 1+3+3²+ …+3^n-1= $\frac{\left(3^n-1\right)}{2}$

For n=1, P(1)=1= $\frac{\left(3^1-1\right)}{2}=\frac{3-1}{2}=\frac{2}{2}=1$

which is true.

Assume that P(k) is true for some positive integer k i.e.,

1+3+3²+ … +3^k–1= $\frac{\left(3^k-1\right)}{2}$

--------(1)

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

Here, 1+3+3²+ … +3^k–1+3^(k+1)–1

$\frac{3^k-1}{2}+3^{k+1-1}$

[By using eq (1)]

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

= $\frac{3^k+2.3^k-1}{2}$

= $\frac{3^k(1+2)-1}{2}$

= $\frac{3^k·3-1}{2}=\frac{3^{k+1}-1}{2}$

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

Hence, from the principle of mathematical induction, the P(n) is true for all natural numbers n.