Mathematical Induction

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Step 1: Base case — For $n=1$: LHS = 1, RHS = $\frac{1(1+1)}{2} = 1$. So, true for $n=1$.<br>Step 2: Inductive hypothesis — Assume true for $n=k$: $1+2+\dots+k = \frac{k(k+1)}{2}$.<br>Step 3: Prove fo…

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Step 1: Base case — $n=1$: LHS = $1^2 = 1$, RHS = $\frac{1(2)(3)}{6} = 1$. True.<br>Step 2: Assume true for $n=k$: $\sum k^2 = \frac{k(k+1)(2k+1)}{6}$.<br>Step 3: For $n=k+1$: Add $(k+1)^2$ to both si…

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Step 1: $n=1$: LHS = $1^3 = 1$, RHS = $\left(\frac{1\cdot2}{2}\right)^2 = 1$. True.<br>Step 2: Assume true for $n=k$.<br>Step 3: For $n=k+1$: Add $(k+1)^3$ to both sides:<br>LHS becomes $\left(\frac{k…

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Step 1: $n=1$: $2^1 = 2 > 1$. True.<br>Step 2: Assume $2^k > k$.<br>Step 3: For $n=k+1$: $2^{k+1} = 2\cdot2^k > 2k$.<br>Since $k \geq 1$, $2k = k + k \geq k + 1$. So $2^{k+1} > k+1$.<br>Thus, $2^{k+1}…

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Step 1: $n=3$: $9 > 8$. True.<br>Step 2: Assume $k^2 > 2(k+1)$ for $k \geq 3$.<br>Step 3: For $n=k+1$: $(k+1)^2 = k^2 + 2k + 1 > 2(k+1) + 2k + 1 = 4k + 3$.<br>Now, $4k + 3 > 2(k+2) = 2k + 4$ since $2k…

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Step 1: $n=1$: LHS = 1, RHS = $1^2 = 1$. True.<br>Step 2: Assume true for $n=k$: $1+3+\dots+(2k-1) = k^2$.<br>Step 3: For $n=k+1$: Add $(2k+1)$ to both sides:<br>LHS = $k^2 + (2k+1) = k^2 + 2k + 1 = (…

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Step 1: $n=1$: $x^1 - y^1 = x - y$, divisible by $x - y$. True.<br>Step 2: Assume $x^k - y^k$ is divisible by $x - y$, so $x^k - y^k = (x - y)P$.<br>Step 3: For $n=k+1$: $x^{k+1} - y^{k+1} = x\cdot x^…

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Step 1: $n=1$: $49 + 16 - 1 = 64$, divisible by 64. True.<br>Step 2: Assume $49^k + 16k - 1 = 64P$.<br>Step 3: For $n=k+1$: $49^{k+1} + 16(k+1) - 1 = 49\cdot49^k + 16k + 15$.<br>Substitute $49^k = 64P…