Prove with induction that $$3\cdot 4 + 5 \cdot7 + 7 \cdot 10 + 9\cdot 13 +\ldots+ (2n + 1)(3n + 1) = \frac{n}{2} \left(4n^2 + 11n + 9 \right)$$
*For n = 1
$(2n + 1)(3n + 1) = n/2 (4n^2 + 11n + 9)$ $(2*1 + 1)(3*1 + 1) = 1/2 (4(1) + 11 + 9)$ $12 = 12$ (proved)
*For n = k
$3*4 + 5*7 + 7*10 + 9*13 +...+ (2k + 1)(3k + 1) = k/2 (4k^2 + 11k + 9)$
*For $n = k + 1$
$3*4 + 5*7 + 7*10 + 9*13 +...+ (2k + 1)(3k + 1) + (2(k + 1) + 1)(3(k + 1) + 1) = (k + 1)/2 (4(k + 1)^2 + 11(k + 1) + 9)$
RHS:
$(k + 1)/2 (4(k + 1)^2 + 11(k + 1) + 9)$
Here's my working:
$3*4 + 5*7 + 7*10 + 9*13 +...+ (2k + 1)(3k + 1) + (2(k + 1) + 1)(3(k + 1) + 1) = (k + 2)/2 (4(k + 1)^2 + 11(k + 1) + 9)$
Since $3*4 + 5*7 + 7*10 + 9*13 +...+ (2k + 1)(3k + 1) = k/2 (4k^2 + 11k + 9)$, so
$k/2 (4k^2 + 11k + 9) + (2k + 3)(3k + 4) = (k + 1)/2 (4(k + 1)^2 + 11(k + 1) + 9)$
LHS:
$k/2 (4k^2 + 11k + 9) + (2k + 3)(3k + 4)$
$k/2 (4k^2 + 11k + 9) + (6k^2 + 17k + 12)$
$k/2 (4k^2 + 11k + 9) + k/2(12k + 34 + 24/k)$
$k/2 (4k^2 + 11k + 9 + 12k + 34 + 24/k)$
$k/2 (4k^2 + 23k + 43 + 24/k)$
Now I'm stucked. Can't solve LHS = RHS. Please help me
You have the right idea but taking out the factor of $\frac{k}{2}$ was misguided since you know that it's actually a factor of $\frac{k+1}{2}$ which is required.
There's nothing for it but to simply multiply out:-
$\frac{k}{2}(4k^2+11k+9)+(2k+3)(3k+4)=\frac{1}{2}(4k^3+23k^2+43k+24)=\frac{1}{2}(k+1)(4k^2+19k+24)$
And you're there.