Further Pure Mathematics 4PM1 / Sequences and Series / 1.271.27: Arithmetic Series and an Inequality4PM1/1R/June/2019 — Question 6 · 10 marksMark as done · Save for later(a) Show that∑r=1n(4r−3)=n(2n−1).\sum_{r=1}^{n}(4r-3)=n(2n-1).r=1∑n(4r−3)=n(2n−1).(3)(b) Hence, or otherwise, find the least value of n\displaystyle nn such that∑r=1n(4r−3)>1000.\sum_{r=1}^{n}(4r-3)>1000.r=1∑n(4r−3)>1000.(3)Given that Sn=n(2n−1)\displaystyle S_n=n(2n-1)Sn=n(2n−1), tn=4n−3\displaystyle t_n=4n-3tn=4n−3 and that18+3tn+7=Sn+4,18+3t_{n+7}=S_{n+4},18+3tn+7=Sn+4,(c) find the value of n\displaystyle nn.(4)▸ Mark scheme← 1.30: 4PM1/1/January/2019 — Question 21.28: 4PM1/2/June/2019 — Question 7 →
