find-the-nth-term-and-the-sum-to-n-termof-the-following-seried-i-4-6-9-13-18-ii-11-23-59-167-

Question Number 40218 by scientist last updated on 17/Jul/18

$${find}\:{the}\:{nth}\:{term}\:{and}\:{the}\:{sum}\:{to}\:{n}\:\:{termof}\:{the}\:{following}\:{seried} \\ $$$$\left({i}\right)\:\mathrm{4}+\mathrm{6}+\mathrm{9}+\mathrm{13}+\mathrm{18}+… \\ $$$$\left({ii}\right)\:\mathrm{11}+\mathrm{23}+\mathrm{59}+\mathrm{167}+… \\ $$

Answered by math1967 last updated on 17/Jul/18

i)4+6+9+13+18+..=Sn 4+6+9+13+...=Sn by subtracting 4+{2+3+4+5+......(n−1)term}=t_n ∴t_n =4+(((n−1)(n+2))/2)=((n^2 +n+6)/2) now S_n =(1/2)(1^2 +2^2 +....n^2 )+(1/2)(1+2+..)+3n =(1/(12))(n+1)(2n+1)+(1/4)n(n+1)+3n

$$\left.{i}\right)\mathrm{4}+\mathrm{6}+\mathrm{9}+\mathrm{13}+\mathrm{18}+..={Sn} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{4}+\mathrm{6}+\mathrm{9}+\mathrm{13}+…={Sn} \\ $$$${by}\:{subtracting}\: \\ $$$$\mathrm{4}+\left\{\mathrm{2}+\mathrm{3}+\mathrm{4}+\mathrm{5}+……\left({n}−\mathrm{1}\right){term}\right\}={t}_{{n}} \\ $$$$\therefore{t}_{{n}} =\mathrm{4}+\frac{\left({n}−\mathrm{1}\right)\left({n}+\mathrm{2}\right)}{\mathrm{2}}=\frac{{n}^{\mathrm{2}} +{n}+\mathrm{6}}{\mathrm{2}} \\ $$$${now}\:{S}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +….{n}^{\mathrm{2}} \right)+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{2}+..\right)+\mathrm{3}{n} \\ $$$$=\frac{\mathrm{1}}{\mathrm{12}}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{4}}{n}\left({n}+\mathrm{1}\right)+\mathrm{3}{n} \\ $$

Commented by scientist last updated on 17/Jul/18

from this solution i guess, suppose (1+2+3+4+...n) =((n(n+1))/2) so where did u now get (n−1)(n+2) instead (n−1)(n)

$${from}\:{this}\:{solution}\:{i}\:{guess},\:\:\:{suppose}\:\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+…{n}\right) \\ $$$$=\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$${so}\:{where}\:{did}\:{u}\:{now}\:{get}\:\left({n}−\mathrm{1}\right)\left({n}+\mathrm{2}\right)\:{instead}\:\:\:\left({n}−\mathrm{1}\right)\left({n}\right) \\ $$$$ \\ $$

Commented by math1967 last updated on 18/Jul/18

4+{2+3+4+...(n−1)term} 4+(((n−1){2×2+(n−1−1)×1})/2) [using S_(n ) =(n/2){2a+(n−1)d}] 4+(((n−1)(4+n−2))/2) 4+(((n−1)(n+2))/2)=((8+n^2 +n−2)/2)=((n^2 +n+6)/2)

$$\mathrm{4}+\left\{\mathrm{2}+\mathrm{3}+\mathrm{4}+…\left({n}−\mathrm{1}\right){term}\right\} \\ $$$$\mathrm{4}+\frac{\left({n}−\mathrm{1}\right)\left\{\mathrm{2}×\mathrm{2}+\left({n}−\mathrm{1}−\mathrm{1}\right)×\mathrm{1}\right\}}{\mathrm{2}} \\ $$$$\left[{using}\:{S}_{{n}\:} =\frac{{n}}{\mathrm{2}}\left\{\mathrm{2}{a}+\left({n}−\mathrm{1}\right){d}\right\}\right] \\ $$$$\mathrm{4}+\frac{\left({n}−\mathrm{1}\right)\left(\mathrm{4}+{n}−\mathrm{2}\right)}{\mathrm{2}} \\ $$$$\mathrm{4}+\frac{\left({n}−\mathrm{1}\right)\left({n}+\mathrm{2}\right)}{\mathrm{2}}=\frac{\mathrm{8}+{n}^{\mathrm{2}} +{n}−\mathrm{2}}{\mathrm{2}}=\frac{{n}^{\mathrm{2}} +{n}+\mathrm{6}}{\mathrm{2}} \\ $$

Commented by scientist last updated on 18/Jul/18

$${Got}\:{it}\:{sir},\:{shukrah} \\ $$

Answered by math1967 last updated on 17/Jul/18

ii)11+23+59+167+... 11+23 +59+.... 11+12(1+3+3^2 +....)=t_n t_n =6×3^(n−1) +5 S_n =6(1+3+3^2 +....)+5n =6×((3^n −1)/(3−1)) +5n=3(3^n −1)+5n

$$\left.{ii}\right)\mathrm{11}+\mathrm{23}+\mathrm{59}+\mathrm{167}+… \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{11}+\mathrm{23}\:+\mathrm{59}+…. \\ $$$$\mathrm{11}+\mathrm{12}\left(\mathrm{1}+\mathrm{3}+\mathrm{3}^{\mathrm{2}} +….\right)={t}_{{n}} \\ $$$${t}_{{n}} =\mathrm{6}×\mathrm{3}^{{n}−\mathrm{1}} +\mathrm{5} \\ $$$${S}_{{n}} =\mathrm{6}\left(\mathrm{1}+\mathrm{3}+\mathrm{3}^{\mathrm{2}} +….\right)+\mathrm{5}{n} \\ $$$$=\mathrm{6}×\frac{\mathrm{3}^{{n}} −\mathrm{1}}{\mathrm{3}−\mathrm{1}}\:+\mathrm{5}{n}=\mathrm{3}\left(\mathrm{3}^{{n}} −\mathrm{1}\right)+\mathrm{5}{n} \\ $$

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