InterviewSolution
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InterviewSolution
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Find the nth term of the series `1+5+18+58+179+"..."`. |
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Answer» The sequence of first consctive differences is `4,13,40,121,"…"` and second consecutive differences is `9,27,81,"….",`. Clearly, it is an GP with comon ratio 3. So, let the nth term and sum of the series upto n terms of the series be `T_(n)` and `S_(n)`, respectively. Then, `S_(n)=1+5+18+58+179+".."+T_(n-1)+T_(n)"....(i)"` `S_(n)=1+5+18+58+".."+T_(n-1)+T_(n)"....(ii)"` Subtracting Eq. (ii) from Eq.(i), we get `0=1+4+13+40+121+".."+(T_(n)-T_(n))-T_(n)` `implies T_(n)=1+4+13+40+121+".." " upto n terms "` or ` T_(n)=1+4+13+40+121+".."+t_(n-1)+t_(n)"....(iii)"` ` T_(n)=1+4+13+40+".."+t_(n-1)+t_(n)"....(iv)"` Now, subtracting Eq. (iv) from Eq. (iii), we get ` 0=1+3+9+27+81+".."+(t_(n)-t_(n-1))-t_(n)` or `t_(n)=1+3+9+27+81+".." " upto n terms "` `(1*(3^(n)-1))/((3-1))=(1)/(2)(3^(n)-1)` ` therefore T_(n)=sumt_(n)=(1)/(2)(sum3^(n)-sum1)` `=(1)/(2){(3+3^(2)+3^(3)+"..."+3^(n))-n}` `=(1)/(2){(3(3^(n)-1))/((3-1))-n}` `=(3)/(4)(3^(n)-1)-(1)/(2)n` |
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