find the sum of the gp :… to n terms
sum of a g.p. series is represented by the formula, , when |r|<1. ‘sn’ represents the sum of the g.p. series upto nth terms, ‘a’ represents the first term, ‘r’ represents the common ratio and ‘n’ represents the number of terms.here,a = 1r = (ratio between the n term and n-1 term) n terms∴ ⇒∴
hint: first of all, multiply and divide 9 in both the series after taking common 5 and 0.6 from them respectively. then make a g.p of the form \[10,{{10}^{2}},{{10}^{3}}....\] by writing 9 = 10 – 1, 99 = 100 – 1, and so on. finally, use the formula for the sum of n terms of g.p that is \[\dfrac{a\left( 1-{{r}^{n}} \right)}{\left( 1-r \right)}\].complete step-by-step solution -in this question, we have to find the sum of the n terms of the series.(i) 5 + 55 + 555 + ….(ii) .6 + .66 + .666 + ….let us consider the first seriess = 5 + 55 + 555 + ….. n termsby taking out 5 common from the above series, we get, s = 5 (1 + 11 + 111 + 1111 …. n terms)by multiplying 9 on both the sides of the above equation, we get, 9s = 5 [9 + 99 + 999 …… n times]\[\dfrac{9s}{5}=9+99+999.....\text{ n times}\]we can also write the above equation as, \[\dfrac{9s}{5}=\left( 10-1 \right)+\left( 100-1 \right)+\left( 1000-1 \right)....\text{ n times}\]by simplifying the above equation and taking the term 1 separately, we get, \[\dfrac{9s}{5}=\left( {{10}^{1}}+{{10}^{2}}+{{10}^{3}}.....\text{ n times} \right)-\left( 1+1+1......\text{ n times} \right)\]we know that 1 added n times is equal to n and the nth term of 1, 2, 3…. n. so, we get, \[\dfrac{9s}{5}=\left( {{10}^{1}}+{{10}^{2}}+{{10}^{3}}{{....10}^{n}} \right)-\left( n \right)\]we know that the series of the form \[a,ar,a{{r}^{2}},a{{r}^{3}}.....a{{r}^{n}}\] is in g.p and sum of the n terms of this g.p is given by \[\dfrac{a\left( 1-{{r}^{n}} \right)}{\left( 1-r \right)}\]in the above equation, we can see that \[10,{{10}^{2}},{{10}^{3}}....\] are in g.p with the first term and common ratio as 10. so, we get, \[\dfrac{9s}{5}=\dfrac{10\left( 1-{{\left( 10 \right)}^{n}} \right)}{1-10}-n\]\[\dfrac{9s}{5}=\dfrac{10\left( 1-{{\left( 10 \right)}^{n}} \right)}{-9}-n\]\[\dfrac{9s}{5}=\dfrac{10}{9}\left( {{10}^{n}}-1 \right)-n\]\[\dfrac{9s}{5}=\dfrac{10\left( {{10}^{n}}-1 \right)-9n}{9}\]by multiplying \[\dfrac{5}{9}\] on both the sides of the above equation, we get, \[s=\dfrac{5}{9}\left[ \dfrac{10\left( {{10}^{n}}-1 \right)-9n}{9} \right]\]\[s=\dfrac{50\left( {{10}^{n}}-1 \right)}{81}-\dfrac{5}{9}n\]hence, the sum of the sequence 5 + 55 + 555….. up to n terms is equal to \[\dfrac{50\left( {{10}^{n}}-1 \right)}{81}-\dfrac{5n}{9}\]let us consider the second set.s = 0.6 + 0.66 + 0.666 + ….. n times.by taking out 0.6 common, we can write the sequence as, s = 6 [0.1 + 0.11 + 0.111 ….. n terms]by multiplying and dividing rhs by 9, we get,\[s=\dfrac{6}{9}\left[ 0.9+0.99+0.999.....n\text{ terms} \right]\]\[s=\dfrac{2}{3}\left[ \dfrac{9}{10}+\dfrac{99}{100}+\dfrac{999}{1000}+\dfrac{9999}{10000}.....n\text{ terms} \right]\]we can write the above equation as, \[s=\dfrac{2}{3}\left[ \dfrac{\left( 10-1 \right)}{10}+\dfrac{\left( 100-1 \right)}{100}+\dfrac{\left( 1000-1 \right)}{1000}+\dfrac{\left( 10000-1 \right)}{10000}.....n\text{ terms} \right]\]\[s=\dfrac{2}{3}\left[ \left( 1-\dfrac{1}{10} \right)+\left( 1-\dfrac{1}{100} \right)+\left( 1-\dfrac{1}{1000} \right)+\left( 1-\dfrac{1}{10000} \right).....n\text{ terms} \right]\]\[s=\dfrac{2}{3}\left[ \left( 1+1+1......n\text{ terms} \right)-\left( \dfrac{1}{10}+\dfrac{1}{100}+\dfrac{1}{1000}.+...n\text{ terms} \right) \right]\]here, we can see that \[\dfrac{1}{10}+\dfrac{1}{{{\left( 10 \right)}^{2}}}+\dfrac{1}{{{\left( 10 \right)}^{3}}}+.....n\text{ terms}\] is in gp with \[a=\dfrac{1}{10}\] and \[r=\dfrac{1}{10}\].we know that the sum of n terms of the g.p is \[\dfrac{a\left( 1-{{r}^{n}} \right)}{\left( 1-r \right)}\]. so, we get, \[s=\dfrac{2}{3}\left[ \left( n \right)-\dfrac{\dfrac{1}{10}\left( 1-{{\left( \dfrac{1}{10} \right)}^{n}} \right)}{1-\dfrac{1}{10}} \right]\]\[s=\dfrac{2}{3}\left[ n-\dfrac{\dfrac{1}{10}\left( 1-\dfrac{1}{{{10}^{n}}} \right)}{\dfrac{9}{10}} \right]\]\[s=\dfrac{2}{3}\left[ n-\dfrac{1}{9}\left( 1-\dfrac{1}{{{10}^{n}}} \right) \right]\]we know that \[\dfrac{1}{{{a}^{n}}}={{a}^{-n}}\]. by using this, we get, \[s=\dfrac{2}{3}\left( n-\dfrac{1}{9}\left( 1-{{10}^{-n}} \right) \right)\]so, \[s=\dfrac{2}{3}n-\dfrac{2}{27}\left[ 1-{{\left( 10 \right)}^{-n}} \right]\]hence, we get the sum of the sequence 0.6 + 0.66 + 0.666 + ….. n terms as \[\dfrac{2}{3}n-\dfrac{2}{27}\left[ 1-{{\left( 10 \right)}^{-n}} \right]\].note: in this question, students must note that whenever we are in need to find the sum of series of the type a + aa + aaa + aaaa ….., we always have to make the series of 9 + 99 + 999 + ….. type by multiplying and dividing the whole series by 9 so that we can get a g.p of the type \[10,{{10}^{2}},{{10}^{3}}\]…. also in this question, some students make this mistake of considering 5 + 55 + 555 + …. or 0.6 + 0.66 + 0.666 + …. in g.p which is wrong because we can see that there is no common ratio in these sequences. write a pair of intergers whose sum gives use

summation of series of natural numbers -


 

- wherein

sum of first n natural numbers



 

 and

 

summation of series of natural numbers -


 

- wherein

sum of  squares of first n natural numbers



 

 

3+7+14+24+37 15k

let tn=an2+bn+c            because difference is in a.p: [4,7,10,13]

at n=1     (i)

    n=2      (ii)

    n=3        (iii)

    (ii)-(i)        (iv)

    (iii)-(ii)       (v)

    (v)-(iv)      

                    

 

from (i) 

from (ii) 







 



      

 

put n=15   

               

               

               

               

 




option 1)

126

this option is incorrect

option 2)

122

this option is correct

option 3)

81

this option is incorrect

option 4)

119

this option is incorrect answer: k=7 [readmore]typed answera) δg° = -rtlnksp δg° = -(8.314 jmol-k)(298k)ln(4.6×10-2 )= 7628.72 j = 7.63 kjb) ∆g° = nfe°e° = ∆g°-nfe° = 7.63-(2)(96,000)(11000)e° =-0.0397 v using the long division method: sum of peak value of load current and capacitor current answer: `4+1+(1)/(4)+(1)/(6)+...`
solution: n/a the sum of a pmf over its entire domain must equal 1.
so cς(x=1 to ∞)σ(y=1 to ∞)[(1/4^x)(1/3^y)] = 1.
 
ς(i=1 to ∞)[1/n^i] = 1/(n-1). so
 
ς(y=1 to ∞)[(1/3^y)] = 1/(3-1) = 1/2 and
ς(x=1 to ∞)[(1/4^x)] = 1/(4-1) =1/3.
 
so, σ(x=1 to ∞)σ(y=1 to ∞)[(1/4^x)(1/3^y)] = (1/2)*(1/3) = 1/6.
 
to make the sum = 1, then c=6. [readmore]typed answer2no + o2 → 2no2rate=knomo2n2nd order in no → m = 23rd order overall → m + n = 3so 2 + n = 3n = 1rate=k[no]2[o2]