12 questions
Find the set of values of aaa for which ∑r=3a(2r+a)>a2\sum_{r=3}^{a}(2r+a)>a^2∑r=3a(2r+a)>a2.
It is given that ∑r=1nr2=n(n+1)(2n+1)6\sum_{r=1}^nr^2=\dfrac{n(n+1)(2n+1)}6∑r=1nr2=6n(n+1)(2n+1).
Given ∑r=1nr2=n6(n+1)(2n+1)\sum_{r=1}^nr^2=\dfrac n6(n+1)(2n+1)∑r=1nr2=6n(n+1)(2n+1), show ∑r=1n(2r+1)(2r+3)=n3(4n2+18n+23)\sum_{r=1}^n(2r+1)(2r+3)=\dfrac n3(4n^2+18n+23)∑r=1n(2r+1)(2r+3)=3n(4n2+18n+23). Hence find the exact sum of the 25th to the 75th term of the series.
(i) Find ∑r=1k[(−12)r+1+ln(r+1)]\sum_{r=1}^k[(-\dfrac12)^{r+1}+\ln(r+1)]∑r=1k[(−21)r+1+ln(r+1)] in terms of kkk. Simplify your answer.
Given ∑r=1ncos2rθ=sin(2n+1)θ−sinθ2sinθ\sum_{r=1}^n\cos2r\theta=\dfrac{\sin(2n+1)\theta-\sin\theta}{2\sin\theta}∑r=1ncos2rθ=2sinθsin(2n+1)θ−sinθ, where 0<θ<π0<\theta<\pi0<θ<π, show that
(a) A sequence u2,u3,u4,…u_2,u_3,u_4,\ldotsu2,u3,u4,… is such that un=ln(n−1n+1)u_n=\ln(\dfrac{n-1}{n+1})un=ln(n+1n−1), n≥2n\ge2n≥2. Find in terms of NNN an expression for SN=u2+u3+⋯+uN−1+uNS_N=u_2+u_3+\cdots+u_{N-1}+u_NSN=u2+u3+⋯+uN−1+uN, leaving your answer as lna−f(N)\ln a-f(N)lna−f(N), where aaa is positive.
Given ∑r=1nr+22r+1r(r+1)=12−1(n+1)2n+1\sum_{r=1}^n\dfrac{r+2}{2^{r+1}r(r+1)}=\dfrac12-\dfrac1{(n+1)2^{n+1}}∑r=1n2r+1r(r+1)r+2=21−(n+1)2n+11, evaluate ∑r=5∞r+22r+1r(r+1)\sum_{r=5}^{\infty}\dfrac{r+2}{2^{r+1}r(r+1)}∑r=5∞2r+1r(r+1)r+2, leaving the answer as p/qp/qp/q where p,qp,qp,q are real.
It is given that ∑r=1nr2+3r+1(r+2)!=32−1(n+1)!−1(n+2)!\sum_{r=1}^n\dfrac{r^2+3r+1}{(r+2)!}=\dfrac32-\dfrac1{(n+1)!}-\dfrac1{(n+2)!}∑r=1n(r+2)!r2+3r+1=23−(n+1)!1−(n+2)!1.
The sum ∑r=1n4(2r+3)(2r+5)\sum_{r=1}^n\dfrac4{(2r+3)(2r+5)}∑r=1n(2r+3)(2r+5)4 is denoted SnS_nSn. Given Sn=25−22n+5S_n=\dfrac25-\dfrac2{2n+5}Sn=52−2n+52,
Given ∑n=1N3nn+1(1−12n)=12(3N+1N+1−3)\sum_{n=1}^N\dfrac{3^n}{n+1}(1-\dfrac1{2n})=\dfrac12(\dfrac{3^{N+1}}{N+1}-3)∑n=1Nn+13n(1−2n1)=21(N+13N+1−3), deduce ∑n=1N3n(2n+1)(n+1)(n+2)\sum_{n=1}^N\dfrac{3^n(2n+1)}{(n+1)(n+2)}∑n=1N(n+1)(n+2)3n(2n+1), leaving the answer as a(3N+2N+2−b)a(\dfrac{3^{N+2}}{N+2}-b)a(N+23N+2−b) where a,ba,ba,b are constants to be determined.
Given ∑r=1250014r−3+4r+1=10001−14\sum_{r=1}^{2500}\dfrac1{\sqrt{4r-3}+\sqrt{4r+1}}=\dfrac{\sqrt{10001}-1}4∑r=125004r−3+4r+11=410001−1, hence show, without a calculator, that ∑r=0250014r+1+4r+5>24\sum_{r=0}^{2500}\dfrac1{\sqrt{4r+1}+\sqrt{4r+5}}>24∑r=025004r+1+4r+51>24.
The root test is a convergence test for ∑r=0∞ar\sum_{r=0}^{\infty}a_r∑r=0∞ar. It converges when limn→∞∣an∣n<1\lim_{n\to\infty}\sqrt[n]{|a_n|}<1limn→∞n∣an∣<1 and diverges when that limit exceeds 1; when it is 1 the test is inconclusive. Use it to explain why ∑r=0∞(−3r3+rx5r3+7)r\sum_{r=0}^{\infty}(\dfrac{-3r^3+rx}{5r^3+7})^r∑r=0∞(5r3+7−3r3+rx)r converges for all values of xxx.