please answer the 21st one!! it is of GP....answer till the very last step
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LetS_n=1+(1+x)+(1+x+x^2)+(1+x+x^2+x^3)+.....tonterms.
Hence,(1-x)S_n=(1-x)+(1-x)(1+x)+(1-x)(1+x+x^2)+(1-x)(1+x+x^2+x^3)+....to n terms
(1-x)S_n=(1-x)+(1-x^2)+(1-x^3)+(1-x^4)+......to n terms
=n-(x+x^2+x^3+x^4+.......) to n terms
=n-(x(1-x^n))/(1-x)
HenceS_n=n/(1-x)-(x(1-x^n))/(1-x)^2
Note that if x>1 this becomes
S_n=(x(x^n-1))/(x-1)^2-n/(x-1)
I hope it helpful for U
Hence,(1-x)S_n=(1-x)+(1-x)(1+x)+(1-x)(1+x+x^2)+(1-x)(1+x+x^2+x^3)+....to n terms
(1-x)S_n=(1-x)+(1-x^2)+(1-x^3)+(1-x^4)+......to n terms
=n-(x+x^2+x^3+x^4+.......) to n terms
=n-(x(1-x^n))/(1-x)
HenceS_n=n/(1-x)-(x(1-x^n))/(1-x)^2
Note that if x>1 this becomes
S_n=(x(x^n-1))/(x-1)^2-n/(x-1)
I hope it helpful for U
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