Q-Prove-by-mathematical-induction-that-r-1-n-4r-5-2n-2-7n-

Question Number 93474 by Rio Michael last updated on 13/May/20

Q . Prove by mathematical induction that

\underset{r = 1}{\sum^{n}} (4 r + 5) = 2 n^{2} + 7 n

Answered by eswar150933 last updated on 13/May/20

Answered by allizzwell23 last updated on 13/May/20

\boldsymbol P_{1} : 4 (1) + 5 = 9 = 2 {(1)}^{2} + 7 (1) (\boldsymbol true)

\boldsymbol Let \boldsymbol n = \boldsymbol k

\underset{r = 1}{\sum^{n}} (4 r + 5) = 2 k^{2} + 7 k

\boldsymbol Assume \boldsymbol n = \boldsymbol k + 1 \boldsymbol is \boldsymbol true

\underset{r = 1}{\sum^{n}} (4 r + 5) + (4 (r + 1) + 5)

= 2 {(k + 1)}^{2} + 7 (k + 1)

\boldsymbol To \boldsymbol prove \boldsymbol the \boldsymbol truth \boldsymbol of \boldsymbol n = \boldsymbol k + 1

\underset{r = 1}{\sum^{n}} (4 r + 5) + (4 (r + 1) + 5)

= 2 k^{2} + 7 k + (4 (k + 1) + 5)

= 2 k^{2} + 7 k + 4 k + 4 + 5

= 2 k^{2} + 4 k + 2 + 7 k + 7

= 2 (k^{2} + 2 k + 1) + 7 (k + 1)

= 2 {(k + 1)}^{2} + 7 (k + 1)

\boldsymbol DONE!

130520

Commented by Rio Michael last updated on 13/May/20

thanks y^{'} all

thanks y^{'} all

Commented by allizzwell23 last updated on 13/May/20

{you}^{'} re welcom

{you}^{'} re welcom