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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
If `S_1,S_2,S_3` are the sums of first n natural numbers, their squares and their cubes respectively then `S_3(1+8S_1)=` |
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Answer» We have `S_(1)=(1+2+3+...+n)rArr S_(1)=(1)/(2)n(n+1),` `S_(2)=(1^(2)+2^(2)+3^(2)+ ...+n^(2))rArr S_(2)=(1)/(6)n(n+1)(2n+1),` `and S_(3)=(1^(2)+2^(3)+3^(3)+ ...+n^(3)) rArr S_(3)=(1)/(4)n^(2)(n+1)^(2)`. `therefore 9S_(2)^(2)=9xx(1)/(36)*n^(2)(n+1)^(2)(2n+1)^(2)=(1)/(4)n^(2)(n+1)^(2)(2n+1)^(2).` And , `S_(3)(1+8S_(1))=(1)/(4)n^(2)(n+1)^(2)*{1+8*(1)/(2)n(n+1)}` ` =(1)/(4)n^(2)(n+1)^(2)(4n^(2)+4n+1)=(1)/(4)n^(2)(n+1)^(2)(2n+1)^(2).` Hence, `9S_(2)^(2)=S_(3)(1+8S_(1)).` |
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| 2. |
Find the sum `(41+42+43+…+100).` |
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Answer» Correct Answer - 4230 Given sum `=(1+2+3+...+40+41+...+100)-(1+2+3+...+40)` `=((1)/(2)xx100xx101)-((1)/(2)xx40xx41)=(5050-820)=4230.` |
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| 3. |
Find the sum `(2+4+6+8+…+100).` |
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Answer» Correct Answer - 2550 We know that `(1+2+3+…+n)=(1)/(2)n(n+1).` `therefore (2+4+6+8+...+100)=2xx(1+2+3+...+50)=(2xx(1)/(2)xx50xx51)=2550.` |
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| 4. |
If `S_k=(1+2+3+...k)/k` then find the value of `S_1^2+S_2^2+....S_n^2` |
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Answer» `S_(k)=((1+2+3+...+k))/(k)=(k(k+1))/(2k)=(k+1)/(2).` `therefore sum_(k=1)^(n)S_(k)^(2)=sum_(k=1)^(n)(1)/(4)(k+1)^(2)=(1)/(4)(sum_(k=1)^(n)k^(2))+(1)/(2)(sum_(k=1)^(n)k)+(1)/(4)(1+1+ ... n " times")` `=(1)/(4)xx(1)/(6)n(n+1)(2n+1)+(1)/(4)n(n+1)+(1)/(4)n` `=(n)/(24)(2n^(2)+9n+13) " " `[on simplification]. |
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| 5. |
Find the sum to n terms of the series :`3xx8+6xx11+9xx14+dotdotdot` |
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Answer» Correct Answer - `3n(n+1)(n+3)` `T_(n)=("nth term of " 3,6,9,...)xx("nth term of "8,11, 14, ...)` `={3+(n-1)xx3}xx{8+(n-1)xx3}=3n(3n+5)=9n^(2)+15n.` `therefore S_(n) =9(sum_(k=1)^(n)k^(2))+15(sum_(k=1)^(n)k)`. |
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| 6. |
Find the sum of the following series to `n`term: `3xx1^2+5xx2^2+7xx3^2+` |
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Answer» Correct Answer - `(n(n+1)(3n^(2)+5n+1))/(6)` `T_(n)=("nth tem of "3,5,7,..) xx ("nth term of "1^(2),2^(2),3^(2),...)` `={3+(n-1)xx2}xx n^(2)=(2n+1)n^(2)=(2n^(3)+n^(2)).` `therefore S_(n)=2(sum_(k=1)^(n)k^(3))+(sum_(k=1)^(n)k^(2)).` |
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| 7. |
The sum of the series `2/3+8/9+(26)/(27)+(80)/(81)+`to `n`terms is`n-1/2(3^(-n)-1)`(b) `n-1/2(1-3^(-n))`(c) `n+1/2(3^n-1)`(d) `n-1/2(3^n-1)` |
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Answer» `2/8+8/9+26/27+80/81+...` `1-1/3+1-1/9+1-1/27+...` `1-1/(3k)` `sum_(k=1)^(n)1-1/(2k)=n-[1/3((1-(1/3)^n)/(1-1/3))]=n-[(1/3)/(2/3)(1-3^(-n))]` `=n-1/2[1-3^(-n)]`. |
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| 8. |
Find the sum to `n`terms of the series: `1/(1. 3)+1/(3. 5)+1/(5. 7)+` |
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Answer» `S=1/(1*3)+1/(3*5)+1/(5*7)+...+n^(th)`term `S=sum_(r=1)^n 1/(r(r+2))` `S=sum_(r=1)^n((r+2-r)/(r(r+2)))` `S=1/2{sum_(r=1)^n(1/r-1/(r+2))}` `S=1/2{1+1/2-1/(n+1)-11/(n+2)}`. |
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| 9. |
Find the sum `{(6)^(3)+(7)^(3)+(8)^(3)+(9)^(3)+(10)^(3)}.` |
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Answer» Correct Answer - 2800 We know that `(1^(3)+2^(3)+3^(3)+ ...+n^(3))={(1)/(2)n(n+1)}^(2).` `therefore " given sum"={1^(3)+2^(3)+3^(3)+4^(3)+5^(3)+6^(3)+...+(10)^(3)}-{1^(3)+2^(3)+3^(3)+4^(3)+5^(3)}` `=((1)/(2)xx10xx11)^(2)-((1)/(2)xx5xx6)^(2)=(55)^(2)-(15)^(2)=(3025-225)=2800.` |
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| 10. |
Sum the following series to `n`terms: `5+7+13+31+85+` |
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Answer» Difference between terms in GP Method of difference `S_n=5+7+13+31+85+...+T_n-(1)` `S_n=0+5+7+13+31+81+...+T_n-(2)` Subtracting equation 2 from 1 `0=5+2+6+18+...+(T_n-T_(n-1))-T_n` `T_n=5+2(3^(n-1)-1)/(3-1)` `T_n=5+(3^(n-1)-1)` `T_n=4+3^(n-1)` `S_n=sum_(k=1)^n T_k=sum_(k=1)^n 4+3^(k-1)` `s_n=4n+(3^n-1)/2` `=1/2(3^n+8n-1)`. |
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| 11. |
Find the sum of n terms of the series `(1)/((2 xx 5))+(1)/((5xx8))+(1)/((8xx11))+... .` |
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Answer» We have `T_(k)=(1)/(("kth term of "2,5,8, ...) xx("kth term of "5,8,11, ...))` `=(1)/({2+(k-1)xx3}xx{5+(k-1)xx3})` `=(1)/((3k-1)(3k+2))=(1)/(3){(1)/((3k-1))-(1)/((3k+2))}`. `therefore T_(k)=(1)/(3){(1)/((3k-1))-(1)/((3k+2))}. " " `...(i) Putting `k=1,2,3, ..., n` successively in (i), we get `T_(1)=(1)/(3)((1)/(2)-(1)/(5))` `T_(2)=(1)/(3)((1)/(5)-(1)/(8))` `T_(3)=(1)/(3)((1)/(8)-(1)/(11))` ... ... ... ... ... ... ... ... `T_(n) =(1)/(3){(1)/((3n-1))-(1)/((3n+2))}.` Adding columnwise, we get `S_(n)=(T_(1)+T_(2)+T_(3)+...+T_(n))` `=(1)/(3)((1)/(2)-(1)/(3n+2))=(n)/(2(3n+2)).` |
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| 12. |
Find the sum of the first n terms of the series : `3 + 7 + 13 +21 + 31 +dot dot dot` |
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Answer» Let `S_(n) =3+7+13+21+31+...+T_(n-1)+T_(n)` `S_(n) =3+7+13+21+...+T_(n-2)+T_(n-1)+T_(n)` On subtraction, we get `0=3+[4+6+8+10+..."to"(n-1)" terms"]-T_(n)` `rArr T_(n)=3+[4+6+8+... "to"(n-1)"terms"]` `=3+((n-1))/(2)xx[2xx4+(n-2)xx2]` `=3+(n-1)(n+2)=(n^(2)+n+1).` `therefore S_(n)=sum_(k=1)^(n)T_(k) =sum_(k=1)^(n)(k^(2)+k+1)` `=sum_(k=1)^(n)k^(2)+sum_(k=1)^(n)k+n " "[because 1+1+1+... "to n terms"=n]` `=(1)/(6)n(n+1)(2n+1)+(1)/(2)n(n+1)+n=(1)/(3)n(n^(2)+3n+5).` |
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| 13. |
Write the sum of `n`term fo a series whose `r^(t h)`term is: `r+2^rdot` |
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Answer» Correct Answer - `(1)/(2)n(n+1)+2^(n+1)-2` Required sum `=sum_(r=1)^(n)T_(r)=sum_(r=1)^(n)(r+2^(r))` `=sum_(r=1)^(n)r+sum_(r=1)^(n)2^(r)=(1)/(2)n(n+1)+(2+2^(2)+2^(3)+...+2^(n))` `=(1)/(2)n(n+1)+(2(2^(n)-1))/((2-1))=(1)/(2)n(n+1)+2^(n+1)-2.` |
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| 14. |
Find the sum of first `n`terms of the following series:`5+11+19+29+41+ dot` |
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Answer» Let `S_(n)=5+11+19+29+...+T_(n-1)+T_(n)` `S_(n)=5+11+19+...+T_(n-2)+T_(n-1)+T_(n)` On substraction, we get `0=5+[6+8+10+12= ..." to "(n-1)" terms"]-T_(n)` `rArr T_(n)=5+((n-1)xx[2xx6+(n-2)xx2])/(2)` `rArr T_(n)=5+(n-1)(n+4)` `rArr T_(n)=n^(2)+3n+1.` `therefore S_(n) =sum_(k=1)^(n)T_(k)=sum_(k=1)^(n)(k^(2)+3k+1)=sum_(k=1)^(n)k^(2)+3sum_(k=1)^(n)k+n` `=(1)/(6)n(n+1)(2n+1)+3xx(1)/(2)n(n+1)+n` `=(1)/(3)n(n+2)(n+4).` |
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| 15. |
Sum the following series to `n`terms:`1/(1. 6)+1/(6. 11)+1/(11. 16)+1/(16. 21)++1/((5n-4)(5n+1))` |
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Answer» Correct Answer - `(n)/((5n+1))` `T_(n)=(1)/(5)xx{(1)/(5n-4)-(1)/(5n+1)}`. `therefore T_(1)=(1)/(5)xx{(1)/(1)-(1)/(6)}` `T_(2)=(1)/(5)xx{(1)/(6)-(1)/(11)}` `T_(3)=(1)/(5)xx{(1)/(11)-(1)/(16)}` ... ... ... ... ... ... ... ... ... ... `T_(n-1)=(1)/(5)xx{(1)/(5n-9)-(1)/(5n-4)}` `T_(n) =(1)/(5) xx {(1)/(5n-4)-(1)/(5n+1)}`. adding columnwise, we get `S_(n)=(T_(1)+T_(2)+ ... +T_(n))=(1)/(5)(1-(1)/(5n+1))=(5n+1-1)/(5(5n+1))=(n)/((5n+1))`. |
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| 16. |
Find the value of `11^2+12^2+13^2++20^2dot` |
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Answer» Correct Answer - 2485 We know that `(1^(2)+2^(2)+3^(2)+ ...+n^(2))=(1)/(6)n(n+1)(2n+1).` `therefore " given sum"={1^(2)+2^(2)+...+(10)^(2)+(11)^(2)+...+(20)^(2)}-{1^(2)+2^(2)+..+(10)^(2)}` `=((1)/(6)xx20xx21xx41)-((1)/(6) xx 10xx11 xx21)=(2870-385)=2485.` |
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| 17. |
Find the 50th term of the series `2+3+6+11+18+…`. |
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Answer» Let `S_(n)=2+3+6+11+18+...+T_(n-1)+T_(n)`. Again `S_(n)=2+3+6+11+...+T_(n-1)+T_(n)`. On substracting, we get `0=2+[1+3+5+7+... "to"(n-1)"terms"]-T_(n)` `rArr T_(n)=2+[1+3+5+7+... "to"(n-1)"terms"]` `=2+((n-1))/(2)*[2xx1+(n-2)xx2]=(n-1)^(2)+2.` `rArr T_(50)={(50-1)^(2)+2}={(49)^(2)+2}=(2401+2)=2403.` |
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| 18. |
Find the sum of the following series to `n`term: `2^3+4^3+6^3+8^3+ ` |
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Answer» Correct Answer - `2n^(2)(n+1)^(2)` `T_(n)={2+(n-1)xx2}^(3)=(2n)^(3)=8n^(3)` `therefore S_(n)=8{sum_(k=1)^(n)k^(3)}=8xx(1)/(4)n^(2)(n+1)^(2)=2n^(2)(n+1)^(2).` |
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| 19. |
Sum the series `3. 8+6. 11+9. 14+ `to `n`terms. |
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Answer» We have `T_(k)=("kth term of "3,6,9, …) xx ("kth term of "8,11,14,…)` `={3+(k-1)xx3}xx[8+(k-1)xx 3} = 3k(3k+5)` `=(9k^(2)+15k).` `S_(n)=sum_(k=1)^(n)T_(k)` `=sum_(k=1)^(n)(9k^(2)+15k)=9(sum_(k=1)^(n)k^(2))+15(sum_(k=1)^(n)k)` `=9*{(1)/(6)n(n+1)(2n+1)}+15*{(1)/(2)n(n+1)}` `=(3)/(2)n(n+1){(2n+1)+5}=3n(n+1)(n+3)`. Hence, the required sum is `3n(n+1)(n+3)`. |
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| 20. |
Find the sum of the series: `(2^(2)+4^(2)+6^(2)+8^(2)+ ..."to n terms")` |
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Answer» Correct Answer - `(2)/(3)n(n+1)(2n+1)` `T_(n)=[2+(n-1)xx2}^(2)=(2n)^(2)=4n^(2).` `therefore S_(n)=sum_(k=1)^(n)T_(k)=4(sum_(k=1)^(n)k^(2))=4xx(1)/(6)n(n+1)(2n+1).` |
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| 21. |
Find the sum of series`(3^3=2^3)+(5^3=4^3)+(7^3=6^3)+`to `n`terms |
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Answer» We have `T_(k)={(2k+1)^(3)-(2k)^(3)}` `={8k^(3)+1+6k(2k+1)-8k^(3)}=(12k^(2)+6k+1).` (i) sum to n terms is given by `S_(n) =12 sum_(k=1)^(n)k^(2)+6sum_(k=1)^(n)k+n` `=12xx(1)/(6)n(n+1)(2n+1)+6xx(1)/(2)n(n+1)+n` `=2n(n+1)(2n+1)+3n(n+1)+n` `=n(n+1)[4n+2+3]+n=n[(n+1)(4n+5)+1]` `=n(4n^(2)+9n+6)=(4n^(3)+9n^(2)+6n).` (ii) Sum to 10 terms is given by `S_(10) =10 xx[4 xx10^(2)+9xx10+6]=4960.` |
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| 22. |
Find the sum of n terms of the series whose nth term is: `n(n+3)` |
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Answer» We have , `t_(k)=k(k+3)=(k^(2)+3k)`. `therefore ` sum to n terms is given by `S_(n)=sum_(k=1)^(n)T_(k)` `=sum_(k=1)^(n)(k^(2)+3k)=sum_(k=1)^(n)k^(2)+3sum_(k=1)^(n)k` `=(1)/(6)n(n+1)(2n+1)+3*(1)/(2)n(n+1)=(1)/(6){n(n+1)(2n+1)+9n(n+1)}` `=(1)/(6)n(n+1)(2n+1+9)=(1)/(6)n(n+1)*2(n+5)=(1)/(3)n(n+1)(n+5).` Hence, the required sum is `(1)/(3)n(n+1)(n+5).` |
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| 23. |
The sum of 10 terms of the series `sqrt2 + sqrt6 + sqrt18 +...` is |
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Answer» Correct Answer - `121(sqrt(6)+sqrt(2))` Given series a geometric series in which `a =sqrt(2),r=sqrt(3) and n=10.` `therefore S=(a(r^(n)-1))/((r-1))=(sqrt(2){(sqrt(3))^(10)-1})/((sqrt(3)-1))xx((sqrt(3)+1))/((sqrt(3)+1))=(sqrt(2)(243-1)(sqrt(3)+1))/(2)` `rArr S=121(sqrt(6)+sqrt(2)).` |
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| 24. |
Find the sum to n terms of the series whose nth term is `n^2+2^n`. |
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Answer» We have, `T_(k)=(k^(2)+2^(k))`. `therefore S_(n)=sum_(k=1)^(n)T_(k)` `=sum_(k=1)^(n)(k^(2)+2^(k))=sum_(k=1)^(n)k^(2)+sum_(k=1)^(n)2^(k)` `=(1)/(6)n(n+1)(2n+1)+(2+2^(2)+2^(3)+ ...+2^(n))[because sum_(k=1)^(n)k^(2)=(1)/(6)n(n+1)(2n+1)]` `=(1)/(6)n(n+1)(2n+1)+(2(2^(n)-1))/((2-1)) " "[because 2+2^(2)+ ... +2^(n)" is a GP"]` `=(1)/(6)n(n+1)(2n+1)+2(2^(n)-1).` Hence, the required sum is `(1)/(6)n(n+1)(2n+1)+2(2^(n)-1).` |
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| 25. |
Find the sum of the series whose nth term is given by: ` (3n^(2)-3n+2)` |
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Answer» Correct Answer - `2n(n^(2)+1)` `T_(n)=(3n^(2)-3n+2).` `therefore S_(n)=3(sum_(k=1)^(n)k^(2))-3(sum_(k=1)^(n)k)+2n " "[because 2+2+2+.. "n times" = 2n]`. |
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| 26. |
Find the sum of n terms of the series whose nth term is: `n^3-3^n` |
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Answer» Correct Answer - `(1)/(4)n^(2)(n+1)^(2)-(3)/(2)(3^(n)-1)` `T_(n)=(n^(3)-3^(n)). ` `because S_(n)=(sum_(k=1)^(n)k^(3))-(sum_(k=1)^(n)3^(k))=(1)/(4)n^(2)(n+1)^(2)-{3+3^(2)+...+3^(n)}` `rArr S_(n) ={(1)/(4)n^(2)(n+1)^(2)-(3(3^(n)-1))/((3-1))}`. |
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| 27. |
Find the sum of the series whose nth term is given by: `(3n^(2)+2n)` |
| Answer» Correct Answer - `(1)/(2)n(n+1)(2n+3)` | |
| 28. |
Find the sum to n term of the series whose nth term is `n(n+1)(n+4)` |
| Answer» Correct Answer - `(1)/(12)n(n+1)(n+2)(3n+17)` | |
| 29. |
Find the sum of the series `2^2+4^2+6^2++(2n)^2` |
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Answer» Correct Answer - `(2)/(3)n(n+1)(2n+1)` `T_(k)=(2k)^(2)=4k^(2).` `therefore S_(n)=sum_(k=1)^(n)T_(k)=4(sum_(k=1)^(n)k^(2))={4xx(1)/(6)n(n+1)(2n+1)}=(2)/(3)n(n+1)(2n+1)`. |
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| 30. |
let `S_(n)` denote the sum of the cubes of the first n natureal numbers and `S_(n)` denote the sum of the fisrt n natural numbers , then `sum_(r=1)^(n)(S_(r))/(S_(4))` equals to |
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Answer» Correct Answer - `(1)/(6)n(n+1)(n+2)` `S_(k)={sum_(r=1)^(k)r^(3)}={(1)/(2)k(k+1)}^(2) and s_(k)={sum_(r=1)^(k)r}=(1)/(2)k(k+1).` `therefore (S_(k))/(s_(k))=({(1)/(2)k(k+1)}^(2))/((1)/(2)k(k+1))=(1)/(2)k(k+1)=(1)/(2)k^(2)+(1)/(2)k` `rArr sum_(k=1)^(n)(S_(k))/(s_(k))=(1)/(2)[sum_(k=1)^(n)k^(2)}+(1)/(2){sum_(k=1)^(n)k}` `=(1)/(2)xx(1)/(6)xxn(n+1)(2n+1)+(1)/(2)xx(1)/(2)n(n+1)` `=(1)/(12){n(n+1)(2n+1)+3n(n+1)}=(1)/(6)n(n+1)(n+2).` |
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| 31. |
If `sum_(k=1)^(n)k=45,` find the value of `sum_(k=1)^(n)k^(3).` |
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Answer» Correct Answer - 2025 Given `(1+2+3+...+n)=45 rArr (1)/(2)n(n+1)=45.` `therefore (sum_(k=1)^(n)k^(3))={(1)/(2)n(n+1)}^(2)=(45)^(2)=2025.` |
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| 32. |
If `sum_(k=1)^(n)k=210,` find the value of `sum_(k=1)^(n)k^(2).` |
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Answer» Correct Answer - 2870 `(1+2+3+...+n)=210rArr (1)/(2)n(n+1)=210rArr n^(2)+n-420=0` `rArr (n+21)(n-20)=0 rArr n=20.` `therefore (1^(2)+2^(2)+3^(2)+...+n^(2))=(1)/(6)n(n+1)(2n+1), " where " n =20` `=((1)/(6)xx20xx21xx41)=2870.` |
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