Question

How do I write (51⋅2)+(52⋅3)+(53⋅4)+...+(5n(n+1))+...in summation notation, and how can I tell if the series converges?

Answer 1

Seems that there is an error in the given series. I think It's actually,

$\displaystyle (51 \cdot 2)+(52 \cdot 3)+(53 \cdot 4)+...+((50+n)(n+1))$

So,

$\displaystyle \begin{aligned}&(51 \cdot 2)+(52 \cdot 3)+(53 \cdot 4)+...+((50+n)(n+1))\\ \\ \Longrightarrow\ \ &((50+1)(1+1))+((50+2)(2+1))+((50+3)(3+1))+...+((50+n)(n+1))\end{aligned}$

Thus we can write this sum in sigma notation as,

$\displaystyle \sum_{k=1}^{n}[(50+k)(k+1)]$

Now we're going to check this sum whether the series is convergent.

$\displaystyle \begin{aligned}&\sum_{k=1}^{n}[(50+k)(k+1)]\\ \\ \Longrightarrow\ \ &\sum_{k=1}^{n}[50k+50+k^2+k]\\ \\ \Longrightarrow\ \ &\sum_{k=1}^{n}[k^2+51k+50]\\ \\ \Longrightarrow\ \ &\sum_{k=1}^{n}k^2+51\sum_{k=1}^{n}k+\sum_{k=1}^{n}50\\ \\ \Longrightarrow\ \ &(1^2+2^2+3^2+...+n^2)+51(1+2+3+...+n)\\ &+(\underbrace{50+50+50+...+50}_n)\\ \\ \Longrightarrow\ \ &\frac{n(n+1)(2n+1)}{6}+51 \cdot \frac{n(n+1)}{2}+50n\end{aligned}$

$\displaystyle \begin{aligned}\\ \\ \Longrightarrow\ \ &\frac{n(n+1)(2n+1)}{6}+\frac{51n(n+1)}{2}+50n\\ \\ \Longrightarrow\ \ &\frac{n(n+1)(2n+1)}{6}+\frac{153n(n+1)}{6}+\frac{300n}{6}\\ \\ \Longrightarrow\ \ &\frac{n(n+1)(2n+1)+153n(n+1)+300n}{6}\\ \\ \Longrightarrow\ \ &\frac{n((n+1)(2n+1)+153(n+1)+300)}{6}\\ \\ \Longrightarrow\ \ &\frac{n((n+1)(2n+1+153)+300)}{6}\\ \\ \Longrightarrow\ \ &\frac{n((n+1)(2n+154)+300)}{6}\\ \\ \Longrightarrow\ \ &\frac{n(2((n+1)(n+77)+150))}{6}\end{aligned}$

$\displaystyle \begin{aligned}\\ \\ \Longrightarrow\ \ &\frac{2n((n+1)(n+77)+150)}{6}\\ \\ \Longrightarrow\ \ &\frac{n((n+1)(n+77)+150)}{3}\\ \\ \Longrightarrow\ \ &\frac{n(n+1)(n+77)}{3}+50n\\ \\ \Longrightarrow\ \ &n\left[\frac{(n+1)(n+77)}{3}+50\right]\end{aligned}$

Here we get a general formula for the sum of the series. So we can say that the series is convergent.