Sum of $n$ terms of the series $frac 1(1 + x)(1 +2x) + frac 1(1 +2x)(1 +3x) + ldots$ is equal to?

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0
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My try : By using Vn Method
enter image description here



But the answer is $frac n(1+x)(1+ (n+1)x)$

Please tell me where I am wrong or what is the correct method to solve it.

Thanks






sequences-and-series







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  • Please rotate photograph.
    – Mohammad Zuhair Khan
    Sep 8 at 5:03







  • 3




    The general term should be $frac1(1+nx)[1+(n+1)x]$ instead of $frac1(1+nx)(1+2nx)$
    – Yuta
    Sep 8 at 5:10














up vote
0
down vote

favorite












My try : By using Vn Method
enter image description here



But the answer is $frac n(1+x)(1+ (n+1)x)$

Please tell me where I am wrong or what is the correct method to solve it.

Thanks






sequences-and-series







share|cite|improve this question























  • Please rotate photograph.
    – Mohammad Zuhair Khan
    Sep 8 at 5:03







  • 3




    The general term should be $frac1(1+nx)[1+(n+1)x]$ instead of $frac1(1+nx)(1+2nx)$
    – Yuta
    Sep 8 at 5:10












up vote
0
down vote

favorite









up vote
0
down vote

favorite











My try : By using Vn Method
enter image description here



But the answer is $frac n(1+x)(1+ (n+1)x)$

Please tell me where I am wrong or what is the correct method to solve it.

Thanks






sequences-and-series







share|cite|improve this question















My try : By using Vn Method
enter image description here



But the answer is $frac n(1+x)(1+ (n+1)x)$

Please tell me where I am wrong or what is the correct method to solve it.

Thanks






sequences-and-series




sequences-and-series






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share|cite|improve this question













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edited Sep 8 at 5:30









Mohammad Zuhair Khan

968422




968422










asked Sep 8 at 5:00









user580093

8815




8815











  • Please rotate photograph.
    – Mohammad Zuhair Khan
    Sep 8 at 5:03







  • 3




    The general term should be $frac1(1+nx)[1+(n+1)x]$ instead of $frac1(1+nx)(1+2nx)$
    – Yuta
    Sep 8 at 5:10
















  • Please rotate photograph.
    – Mohammad Zuhair Khan
    Sep 8 at 5:03







  • 3




    The general term should be $frac1(1+nx)[1+(n+1)x]$ instead of $frac1(1+nx)(1+2nx)$
    – Yuta
    Sep 8 at 5:10















Please rotate photograph.
– Mohammad Zuhair Khan
Sep 8 at 5:03





Please rotate photograph.
– Mohammad Zuhair Khan
Sep 8 at 5:03





3




3




The general term should be $frac1(1+nx)[1+(n+1)x]$ instead of $frac1(1+nx)(1+2nx)$
– Yuta
Sep 8 at 5:10




The general term should be $frac1(1+nx)[1+(n+1)x]$ instead of $frac1(1+nx)(1+2nx)$
– Yuta
Sep 8 at 5:10










1 Answer
1






active

oldest

votes

















up vote
2
down vote



accepted










I am sorry but, being almost blind, I am unable to read your photograph.



Considering $$a_n=frac n(1+x)(1+ (n+1)x)$$ use patrial fraction decomposition to get
$$a_n=fracn+1(n+1) x+1-fracnn x+1$$ which beautifully telescopes.



I am sure that you can take it from here.






share|cite|improve this answer




















  • Yeah , thanks understood the concept , sorry for the bad picture
    – user580093
    Sep 8 at 5:35










  • @user580093. You are welcome ! On this site, if I may suggest, use MathJax for typing; it is quite easy and you will probably get more answers than when posting photographs. Cheers.
    – Claude Leibovici
    Sep 8 at 5:46










Your Answer




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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










I am sorry but, being almost blind, I am unable to read your photograph.



Considering $$a_n=frac n(1+x)(1+ (n+1)x)$$ use patrial fraction decomposition to get
$$a_n=fracn+1(n+1) x+1-fracnn x+1$$ which beautifully telescopes.



I am sure that you can take it from here.






share|cite|improve this answer




















  • Yeah , thanks understood the concept , sorry for the bad picture
    – user580093
    Sep 8 at 5:35










  • @user580093. You are welcome ! On this site, if I may suggest, use MathJax for typing; it is quite easy and you will probably get more answers than when posting photographs. Cheers.
    – Claude Leibovici
    Sep 8 at 5:46














up vote
2
down vote



accepted










I am sorry but, being almost blind, I am unable to read your photograph.



Considering $$a_n=frac n(1+x)(1+ (n+1)x)$$ use patrial fraction decomposition to get
$$a_n=fracn+1(n+1) x+1-fracnn x+1$$ which beautifully telescopes.



I am sure that you can take it from here.






share|cite|improve this answer




















  • Yeah , thanks understood the concept , sorry for the bad picture
    – user580093
    Sep 8 at 5:35










  • @user580093. You are welcome ! On this site, if I may suggest, use MathJax for typing; it is quite easy and you will probably get more answers than when posting photographs. Cheers.
    – Claude Leibovici
    Sep 8 at 5:46












up vote
2
down vote



accepted







up vote
2
down vote



accepted






I am sorry but, being almost blind, I am unable to read your photograph.



Considering $$a_n=frac n(1+x)(1+ (n+1)x)$$ use patrial fraction decomposition to get
$$a_n=fracn+1(n+1) x+1-fracnn x+1$$ which beautifully telescopes.



I am sure that you can take it from here.






share|cite|improve this answer












I am sorry but, being almost blind, I am unable to read your photograph.



Considering $$a_n=frac n(1+x)(1+ (n+1)x)$$ use patrial fraction decomposition to get
$$a_n=fracn+1(n+1) x+1-fracnn x+1$$ which beautifully telescopes.



I am sure that you can take it from here.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Sep 8 at 5:34









Claude Leibovici

113k1155127




113k1155127











  • Yeah , thanks understood the concept , sorry for the bad picture
    – user580093
    Sep 8 at 5:35










  • @user580093. You are welcome ! On this site, if I may suggest, use MathJax for typing; it is quite easy and you will probably get more answers than when posting photographs. Cheers.
    – Claude Leibovici
    Sep 8 at 5:46
















  • Yeah , thanks understood the concept , sorry for the bad picture
    – user580093
    Sep 8 at 5:35










  • @user580093. You are welcome ! On this site, if I may suggest, use MathJax for typing; it is quite easy and you will probably get more answers than when posting photographs. Cheers.
    – Claude Leibovici
    Sep 8 at 5:46















Yeah , thanks understood the concept , sorry for the bad picture
– user580093
Sep 8 at 5:35




Yeah , thanks understood the concept , sorry for the bad picture
– user580093
Sep 8 at 5:35












@user580093. You are welcome ! On this site, if I may suggest, use MathJax for typing; it is quite easy and you will probably get more answers than when posting photographs. Cheers.
– Claude Leibovici
Sep 8 at 5:46




@user580093. You are welcome ! On this site, if I may suggest, use MathJax for typing; it is quite easy and you will probably get more answers than when posting photographs. Cheers.
– Claude Leibovici
Sep 8 at 5:46

















 

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