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Learn/ 11th Class Mathematics/ Unit 6 /Numericals

Unit 6: Sequences and Series — Exercise 6 10

11th Class Mathematics · Unit 6: Sequences and Series

1.(i) Sum the series 1×3 + 2×5 + 3×7 + ... to n terms (ii) Sum the series 1×5 + 2×8 + 3×11 + ... to n terms (iii) Sum the series 1×2 + 2×5 + 3×8 + ... to n terms (iv) Sum the series 1×3×5 + 2×4×6 + 3×5×7 + ... to n terms (v) Sum the series 1×2×4 + 2×3×7 + 3×4×10 + ... to n terms (vi) Sum the series 2² + 4² + 6² + ... to n terms (vii) Sum the series 3² + 6² + 9² + ... to n terms (viii) Sum the series 4×1² + 7×2² + 10×3² + ... to n terms (ix) Sum the series 3 + (3+7) + (3+7+11) + ... to n terms
Given
(i) General term Tk = k(2k+1) = 2k2 + k
Formula
Sum to n terms Sn = sumk=1n(2k2 + k) = 2sum k2 + sum k
= 2left[n(n+1)(2n+1)6right] + n(n+1)2
= n(n+1)(2n+1)3 + n(n+1)2
= n(n+1)left[2n+13 + 12right]
= n(n+1)left[4n+2+36right]
Result
Sum Sn = n(n+1)(4n+5)6
Given
(ii) General term Tk = k(3k+2) = 3k2 + 2k
Formula
Sum to n terms Sn = sumk=1n(3k2 + 2k) = 3sum k2 + 2sum k
= 3left[n(n+1)(2n+1)6right] + 2left[n(n+1)2right]
= n(n+1)(2n+1)2 + n(n+1)
= n(n+1)2[2n+1+2]
Result
Sum Sn = n(n+1)(2n+3)2
Given
(iii) General term Tk = k(3k-1) = 3k2 - k
Formula
Sum to n terms Sn = sumk=1n(3k2 - k) = 3sum k2 - sum k
= 3left[n(n+1)(2n+1)6right] - n(n+1)2
= n(n+1)(2n+1)2 - n(n+1)2
= n(n+1)2[2n+1-1]
Result
Sum Sn = n2(n+1)
Given
(iv) General term Tk = k(k+2)(k+4) = k3 + 6k2 + 8k
Formula
Sum to n terms Sn = sumk=1n(k3 + 6k2 + 8k)
= sum k3 + 6sum k2 + 8sum k
= left[n(n+1)2right]2 + 6left[n(n+1)(2n+1)6right] + 8left[n(n+1)2right]
= n(n+1)left[n(n+1)4 + (2n+1) + 4right]
= n(n+1)left[n(n+1)4 + 2n + 5right]
Result
Sum Sn = n(n+1)(n2 + 9n + 20)4
Given
(v) General term Tk = k(k+1)(3k+1) = 3k3 + 4k2 + k
Formula
Sum to n terms Sn = sumk=1n(3k3 + 4k2 + k)
= 3sum k3 + 4sum k2 + sum k
= 3left[n(n+1)2right]2 + 4left[n(n+1)(2n+1)6right] + n(n+1)2
= n(n+1)left[3n(n+1)4 + 2(2n+1)3 + 12right]
= n(n+1)12[9n(n+1) + 8(2n+1) + 6]
Result
Sum Sn = n(n+1)(9n2 + 25n + 14)12
Given
(vi) General term Tk = (2k)2 = 4k2
Formula
Sum to n terms Sn = sumk=1n 4k2 = 4sum k2
= 4 · n(n+1)(2n+1)6
Result
Sum Sn = 2n(n+1)(2n+1)3
Given
(vii) General term Tk = (3k)2 = 9k2
Formula
Sum to n terms Sn = sumk=1n 9k2 = 9sum k2
= 9 · n(n+1)(2n+1)6
Result
Sum Sn = 3n(n+1)(2n+1)2
Given
(viii) General term Tk = [4+3(k-1)]k2 = (3k+1)k2 = 3k3 + k2
Formula
Sum to n terms Sn = sumk=1n(3k3 + k2) = 3sum k3 + sum k2
= 3left[n(n+1)2right]2 + n(n+1)(2n+1)6
Result
Sum Sn = n(n+1)(9n+13n+2)12
Given
(ix) Pattern Tn = 3 + 7 + 11 + ... + n terms (AP with a=3, d=4)
Formula
nth term is AP sum Tn = n2[2a + (n-1)d] = n2[6 + (n-1)4]
= n2[6 + 4n - 4] = n2[4n + 2] = n(2n+1)
Tk = 2k2 + k
Sum to n terms Sn = sumk=1n(2k2 + k) = 2sum k2 + sum k
= n(n+1)(4n+5)6
2.(i) Sum the series 1² - 2² + 3² - 4² + ... + (2n-1)² - (2n)² (ii) Sum the series where the nth term is T_n = (1² + 2² + ... + n²)/n, to n terms
Given
(i) General term Tn = (2n-1)2 - (2n)2
= 4n2 - 4n + 1 - 4n2 = -4n + 1
Formula
Sum up to n terms Sn = sumk=1n(-4k + 1) = -4sum k + sum 1
= -4 · n(n+1)2 + n = -2n(n+1) + n
= -2n2 - 2n + n = -2n2 - n
Result
Sum Sn = -n(2n+1)
Given
(ii) nth term formula Tn = 12 + 22 + ... + n2n
= n(n+1)(2n+1)/6n = (n+1)(2n+1)6
= 2n2 + 3n + 16
Tk = 13k2 + 12k + 16
Formula
Sum to n terms Sn = sumk=1n Tk = 13sum k2 + 12sum k + 16sum 1
= 13 · n(n+1)(2n+1)6 + 12 · n(n+1)2 + n6
= n(n+1)(2n+1)18 + n(n+1)4 + n6
Result
Sum Sn = n(4n2 + 15n + 17)36
3.(i) Find the sum to n terms of the series whose nth term is 5n² + 2n + 3 (ii) Find the sum to n terms of the series whose nth term is n² + 2n - 3
Given
(i) nth term Tn = 5n2 + 2n + 3
Tk = 5k2 + 2k + 3
Formula
Sum to n terms Sn = sumk=1n(5k2 + 2k + 3) = 5sum k2 + 2sum k + 3sum 1
= 5 · n(n+1)(2n+1)6 + 2 · n(n+1)2 + 3n
= nleft[5(n+1)(2n+1)6 + (n+1) + 3right]
= nleft[5(2n2+3n+1) + 6(n+1) + 186right]
= nleft[10n2 + 15n + 5 + 6n + 6 + 186right]
Result
Sum Sn = n(10n2 + 21n + 29)6
Given
(ii) nth term Tn = n2 + 2n - 3
Tk = k2 + 2k - 3
Formula
Sum to n terms Sn = sumk=1n(k2 + 2k - 3) = sum k2 + 2sum k - 3sum 1
= n(n+1)(2n+1)6 + 2 · n(n+1)2 - 3n
= nleft[(n+1)(2n+1)6 + (n+1) - 3right]
= nleft[(n+1)(2n+1) + 6(n+1) - 186right]
= nleft[2n2 + 3n + 1 + 6n + 6 - 186right]
Result
Sum Sn = n(2n2 + 9n - 11)6
4.(i) Given nth term is 3n² + 5n + 2, find the sum to 2n terms (ii) Given nth term is n² + n - 2, find the sum to 2n terms
Given
(i) nth term Tn = 3n2 + 5n + 2
Tk = 3k2 + 5k + 2
Formula
Sum of first n terms Sn = sumk=1n(3k2 + 5k + 2) = 3sum k2 + 5sum k + 2sum 1
= 3 · n(n+1)(2n+1)6 + 5 · n(n+1)2 + 2n
= n(n+1)(2n+1)2 + 5n(n+1)2 + 2n
= n2[(n+1)(2n+1) + 5(n+1) + 4]
= n2[2n2 + 3n + 1 + 5n + 5 + 4]
= n2[2n2 + 8n + 10] = n(n2 + 4n + 5)
Sum to 2n terms S2n = 2n[(2n)2 + 4(2n) + 5]
= 2n[4n2 + 8n + 5]
Result
Sum to 2n terms S2n = 2n(4n2 + 8n + 5)
Given
(ii) nth term Tn = n2 + n - 2
Tk = k2 + k - 2
Formula
Sum of first n terms Sn = sumk=1n(k2 + k - 2) = sum k2 + sum k - 2sum 1
= n(n+1)(2n+1)6 + n(n+1)2 - 2n
= nleft[(n+1)(2n+1)6 + n+12 - 2right]
= n6[(n+1)(2n+1) + 3(n+1) - 12]
= n6[2n2 + 3n + 1 + 3n + 3 - 12]
= n6[2n2 + 6n - 8] = n3(n2 + 3n - 4) = n(n+4)(n-1)3
Sum to 2n terms S2n = 2n[(2n)+4][(2n)-1]3
= 2n(2n+4)(2n-1)3 = 4n(n+2)(2n-1)3
Result
Sum to 2n terms S2n = 4n(n+2)(2n-1)3