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JEE Main 202227 JulyEvening Shift
Exam
JEE Main
Year
2022
Date
27 July
Shift
Evening Shift
Subject
Mathematics
Chapter
Sequences and Series

JEE Main 2022 · 27 July · Evening Shift · Mathematics · Sequences and Series

2 ^ 3 - 1 ^ 3 1 × 7 + 4 ^ 3 - 3 ^ 3 + 2 ^ 3 - 1 ^ 3 2 × 11 + 6 ^ 3...

23−131×7+43−33+23−132×11+63−53+43−33+23−133×15+…..+303−293+283−273+…+23−1315×63\dfrac{{2}^{3} - {1}^{3}}{1 \times 7} + \dfrac{{4}^{3} - {3}^{3} + {2}^{3} - {1}^{3}}{2 \times 11} + \dfrac{{6}^{3} - {5}^{3} + {4}^{3} - {3}^{3} + {2}^{3} - {1}^{3}}{3 \times 15} + \ldots . . + \dfrac{{30}^{3} - {29}^{3} + {28}^{3} - {27}^{3} + \ldots + {2}^{3} - {1}^{3}}{15 \times 63} is equal to ______.

Correct answer

0

Step-by-step solution

Given series, 23−131×7+43−33+23−132×11+63−53+43−33+23−133×15+…..+303−293+283−273+…+23−1315×63\dfrac{{2}^{3} - {1}^{3}}{1 \times 7} + \dfrac{{4}^{3} - {3}^{3} + {2}^{3} - {1}^{3}}{2 \times 11} + \dfrac{{6}^{3} - {5}^{3} + {4}^{3} - {3}^{3} + {2}^{3} - {1}^{3}}{3 \times 15} + \ldots . . + \dfrac{{30}^{3} - {29}^{3} + {28}^{3} - {27}^{3} + \ldots + {2}^{3} - {1}^{3}}{15 \times 63}

Now finding Tn=∑k=1n[(2k)3−(2k−1)3]n(4n+3){T}_{n} = \dfrac{{∑}_{k = 1}^{n}\left[{\left(2k\right)}^{3} - {\left(2k - 1\right)}^{3}\right]}{n\left(4n + 3\right)}

=∑k=1n4k2+(2k−1)2+2k(2k−1)n(4n+3)= \dfrac{{∑}_{k = 1}^{n}4{k}^{2} + {\left(2k - 1\right)}^{2} + 2k\left(2k - 1\right)}{n\left(4n + 3\right)}

=∑k=1n(12k2−6k+1)n(4n+3)= \dfrac{{∑}_{k = 1}^{n}\left(12{k}^{2} - 6k + 1\right)}{n\left(4n + 3\right)}

=2n(2n2+3n+1)−3n2−3n+nn(4n+3)= \dfrac{2n\left(2{n}^{2} + 3n + 1\right) - 3{n}^{2} - 3n + n}{n\left(4n + 3\right)}

=n2(4n+3)n(4n+3)=n= \dfrac{{n}^{2}\left(4n + 3\right)}{n\left(4n + 3\right)} = n

So,Tn=n{T}_{n} = n

Now finding Sn=∑n=115Tn=15×162=120{S}_{n} = \mathop{∑}_{n = 1}^{15}{T}_{n} = \dfrac{15 \times 16}{2} = 120

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