A) \[\frac{1}{1-x}\] done clear
B) \[\frac{1}{1+x}\] done clear
C) \[\frac{1}{{{(1+x)}^{2}}}\] done clear
D) \[\frac{1}{{{(1-x)}^{2}}}\] done clear
View Solution play_arrowA) 3 done clear
B) 6 done clear
C) 9 done clear
D) 12 done clear
View Solution play_arrowA) \[\frac{3}{16}\] done clear
B) \[\frac{35}{8}\] done clear
C) \[\frac{35}{4}\] done clear
D) \[\frac{35}{16}\] done clear
View Solution play_arrowquestion_answer4) The sum of the series \[1+3x+6{{x}^{2}}+10{{x}^{3}}+........\infty \] will be
A) \[\frac{1}{{{(1-x)}^{2}}}\] done clear
B) \[\frac{1}{1-x}\] done clear
C) \[\frac{1}{{{(1+x)}^{2}}}\] done clear
D) \[\frac{1}{{{(1-x)}^{3}}}\] done clear
View Solution play_arrowquestion_answer5) \[1+3+7+15+31+..........\]to \[n\] terms = [IIT 1963]
A) \[{{2}^{n+1}}-n\] done clear
B) \[{{2}^{n+1}}-n-2\] done clear
C) \[{{2}^{n}}-n-2\] done clear
D) None of these done clear
View Solution play_arrowquestion_answer6) \[2+4+7+11+16+......\]to \[n\] terms = [Roorkee 1977]
A) \[\frac{1}{6}({{n}^{2}}+3n+8)\] done clear
B) \[\frac{n}{6}({{n}^{2}}+3n+8)\] done clear
C) \[\frac{1}{6}({{n}^{2}}-3n+8)\] done clear
D) \[\frac{n}{6}({{n}^{2}}-3n+8)\] done clear
View Solution play_arrowquestion_answer7) Sum of n terms of series \[12+16+24+40+.....\] will be [UPSEAT 1999]
A) \[2\,({{2}^{n}}-1)+8n\] done clear
B) \[2({{2}^{n}}-1)+6n\] done clear
C) \[3({{2}^{n}}-1)+8n\] done clear
D) \[4({{2}^{n}}-1)+8n\] done clear
View Solution play_arrowquestion_answer8) If \[a,\ b,\ c\] are in H.P., then which one of the following is true [MNR 1985]
A) \[\frac{1}{b-a}+\frac{1}{b-c}=\frac{1}{b}\] done clear
B) \[\frac{ac}{a+c}=b\] done clear
C) \[\frac{b+a}{b-a}+\frac{b+c}{b-c}=1\] done clear
D) None of these done clear
View Solution play_arrowA) 1 done clear
B) 0 done clear
C) \[\infty \] done clear
D) 4 done clear
View Solution play_arrowquestion_answer10) \[{{n}^{th}}\] term of the series \[2+4+7+11+.......\]will be [Roorkee 1977]
A) \[\frac{{{n}^{2}}+n+1}{2}\] done clear
B) \[{{n}^{2}}+n+2\] done clear
C) \[\frac{{{n}^{2}}+n+2}{2}\] done clear
D) \[\frac{{{n}^{2}}+2n+2}{2}\] done clear
View Solution play_arrowquestion_answer11) The sum of the series \[1+2x+3{{x}^{2}}+4{{x}^{3}}+.........\]upto \[n\] terms is
A) \[\frac{1-(n+1){{x}^{n}}+n{{x}^{n+1}}}{{{(1-x)}^{2}}}\] done clear
B) \[\frac{1-{{x}^{n}}}{1-x}\] done clear
C) \[{{x}^{n+1}}\] done clear
D) None of these done clear
View Solution play_arrowA) \[{{2}^{n}}-n-1\] done clear
B) \[1-{{2}^{-n}}\] done clear
C) \[n+{{2}^{-n}}-1\] done clear
D) \[{{2}^{n}}-1\] done clear
View Solution play_arrowA) \[\frac{25}{16}-\frac{4n+5}{16\times {{5}^{n-1}}}\] done clear
B) \[\frac{3}{4}-\frac{2n+5}{16\times {{5}^{n+1}}}\] done clear
C) \[\frac{3}{7}-\frac{3n+5}{16\times {{5}^{n-1}}}\] done clear
D) \[\frac{1}{2}-\frac{5n+1}{3\times {{5}^{n+2}}}\] done clear
View Solution play_arrowA) 1 done clear
B) 2 done clear
C) \[\frac{3}{2}\] done clear
D) \[\frac{5}{2}\] done clear
View Solution play_arrowquestion_answer15) The sum of \[i-2-3i+4+.........\]upto 100 terms, where \[i=\sqrt{-1}\] is
A) \[50(1-i)\] done clear
B) \[25i\] done clear
C) \[25(1+i)\] done clear
D) \[100(1-i)\] done clear
View Solution play_arrowquestion_answer16) \[{{99}^{th}}\] term of the series \[2+7+14+23+34+.....\] is [Pb. CET 2003]
A) 9998 done clear
B) 9999 done clear
C) 10000 done clear
D) 100000 done clear
View Solution play_arrowA) 62525 done clear
B) 25625 done clear
C) 62500 done clear
D) None of these done clear
View Solution play_arrow
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