Using the principle of mathematical induction prove that $2 + 4 + 6 + . . . . + 2 n = n^{2} + n$ .

Open in App

Solution

$\begin{matrix} 2 + 4 + 6 + . . . . + 2 n = n^{2} + n l e t n = 1 ∴ 2 \times 1 = 1^{2} + 1 2 = 2 (t r u e) L e t g i v e n e q u a t i o n b e t r u e f o r n = k ∴ 2 + 4 + 6 + . . . . + 2 k = k^{2} + k (1) I n d u c t i o n s t e p l e t n = k + 1 ∴ 2 + 4 + 6 + . . . + 2 (k + 1) = {(k + 1)}^{2} + (k + 1) 2 + 4 + 6 + . . . . . + 2 k + 2 = k^{2} + 1 + 2 k + k + 1 = k^{2} + k + 2 k + 2 2 (1 + 2 + 3 + . . . + (k + 1)) = k^{2} + 3 k + 2 2 [\frac{k + 1}{2} (k + 1 + 1)] \Rightarrow (k + 1) (k + 2) = k^{2} + 3 k + 2 \Rightarrow k^{2} + 3 k + 2 = k^{2} + 3 k + 2 H e n c e p r o v e d \end{matrix}$

Suggest Corrections

Similar questions

2 + 4 + 6 + . . . . . + 2 n = n^{2} + n

Using the principle of mathematical induction, prove that

$1.2.3 + 2.3.4 + 3.4.5 + . . . + n (n + 1) (n + 2) = \frac{n (n + 1) (n + 2) (n + 3)}{4},$ for all $n \in N .$

1^{2} + 2^{2} + 3^{2} + . . . . + n^{2} > \frac{n^{2}}{3}

n \in N

\forall n ϵ N

I_{n} = \int_{0}^{π / 2} c o s^{n} x s i n n x d x

= \frac{1}{2^{n + 1}} [2 + \frac{2^{2}}{2} + \frac{2^{3}}{3} . . . . + \frac{2^{n}}{n}]

Using principle of mathematical induction, prove that $41^{n} - 14^{n}$ is a multiple of 27.

Or
Prove by the principle of mathematical induction $n < 2^{n}$ for all $n ϵ N$ .

Join BYJU'S Learning Program