Mathematics - Mathematical Induction Question with Solution | TestHub

When $\mathrm{k}=1,\mathrm{LHS}=1$ but $\mathrm{RHS}=1+10=11$ $\therefore \mathrm{T}(1)$ is not true Let $\mathrm{T}(k)$ is true. That is $\begin{array}{l}1+3+5+\dots .+(2k-1)=k^2+10\\ \text{ Now, }1+3+5+\dots ..+(2k-1)+(2k+1)\\ =k^2+10+2k+1=(k+1{)}^2+10\end{array}$ $\therefore \mathrm{T}(k+1)$ is true. That is $\mathrm{T}(k)$ is true $\Rightarrow \mathrm{T}(k+1)$ is true. But $\mathrm{T}(n)$ is not true for all $n\in \mathbf{N},$ as $\mathrm{T}(1)$ is not true.