Q.Using mathematical induction, show that $\dfrac{1}{1.4} + \dfrac{1}{4.7} + \dfrac{1}{7.10} + \cdots$ upto n terms $= \dfrac{n}{3n+1}\ \forall\ n \in N$.

Verify the base case $n=1$, then show that if the formula holds for $n=m$, adding the next term makes it hold for $n=m+1$ as well — the algebra collapses via factoring $3m^2+4m+1=(3m+1)(m+1)$.

Given: Show $S(n): \dfrac{1}{1\cdot4}+\dfrac{1}{4\cdot7}+\dfrac{1}{7\cdot10}+\cdots$ to $n$ terms $=\dfrac{n}{3n+1}$, $\forall n\in N$.

The $k$-th term is $\dfrac{1}{(3k-2)(3k+1)}$ (denominators $1,4,7,10,\ldots$ follow $3k-2$).

Step 1. Base case ($n=1$):

$\text{LHS} = \dfrac{1}{1\cdot4} = \dfrac14, \qquad \text{RHS} = \dfrac{1}{3(1)+1} = \dfrac14$

LHS $=$ RHS, so $S(1)$ is true.

Step 2. Inductive hypothesis. Assume $S(m)$ is true for some $m\in N$:

$\dfrac{1}{1\cdot4}+\cdots+\dfrac{1}{(3m-2)(3m+1)} = \dfrac{m}{3m+1}$

Step 3. Inductive step. Add the $(m+1)$-th term, $\dfrac{1}{(3m+1)(3m+4)}$, to both sides:

$\dfrac{m}{3m+1} + \dfrac{1}{(3m+1)(3m+4)} = \dfrac{m(3m+4)+1}{(3m+1)(3m+4)} = \dfrac{3m^2+4m+1}{(3m+1)(3m+4)}$

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