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Question Number 205627 by lmcp1203 last updated on 25/Mar/24

Answered by Rasheed.Sindhi last updated on 27/Mar/24

$$\bullet \Bumpeq \bullet \Bumpeq \bullet \Bumpeq \bullet \Bumpeq \bullet \Bumpeq \bullet \Bumpeq \bullet \Bumpeq \bullet \Bumpeq \bullet \Bumpeq \bullet \Bumpeq \bullet \Bumpeq \bullet \Bumpeq \bullet \Bumpeq \bullet \Bumpeq \bullet \Bumpeq \bullet \Bumpeq \bullet \Bumpeq \bullet \Bumpeq \bullet \Bumpeq \bullet \Bumpeq \bullet$$$$2,6,12,20,30,...,i^2+i$$$$S=\underset{i=1}{\sum ^n}(i^2+i)=\underset{i=1}{\sum ^n}{i}^2+\underset{i=1}{\sum ^n}i$$$$S=\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}$$$$=\frac{n(n+1)(2n+1)+3n(n+1)}{6}$$$$=\frac{n(n+1)(2n+4)}{6}$$$$=\frac{\cancel{2}n(n+1)(n+2)}{\cancel{\underset{3}{6}}}$$$$=\frac{n(n+1)(n+2)}{3}$$$$Lastterm-firstterm=\frac{4}{15}S$$$$(n^2+n)-2=\frac{4}{15}\cdot \frac{\left(n(n+1)(n+2)\right)}{3}$$$$45\stackrel{\times }{(n+2)}(n-1)=4n(n+1)\stackrel{\times }{(n+2)}$$$$45(n-1)=4n(n+1)$$$$4n^2-41n+45=0$$$$(n-9)(4n-5)=0$$$$n=9✓,5/4\notin \mathbb{N}\left(invalid\right)$$$$⋓⋒⋓⋒⋓⋒⋓⋒⋓⋒⋓⋒⋓⋒⋓⋒⋓⋒⋓⋒⋓⋒⋓⋒⋓⋒⋓$$

Commented by lmcp1203 last updated on 27/Mar/24

$$thanks$$

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