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JEE • MATHEMATICS • SEQUENCE OF SERIES

Sequence of series practice · Hard

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Showing 31-37 of 37 questions

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31 Hard
31 of 37
Analyzing +4 | -1
For x∈R,x≠−1x \in R, x \neq-1, if (1+x)2016+x(1+x)2015+x2(1+x)^{2016}+x(1+x)^{2015}+x^{2} (1+x)2014+….+x2016=∑i=02016aixi(1+x)^{2014}+\ldots .+\mathrm{x}^{2016}=\sum_{\mathrm{i}=0}^{2016} \mathrm{a}_{\mathrm{i}} \mathrm{x}^{\mathrm{i}}, then a17\mathrm{a}_{17} is equal to ::
View answer and explanation
Correct option: A
2017!17!2000!\frac{2017 !}{17 ! 2000 !}
Option A (2017!17!2000!\frac{{2017 !}}{{17 ! 2000 !}}) is correct because by simplifying the given expression, we can find the coefficient of x17.
32 Hard
32 of 37
Analyzing +4 | -1
Suppose a,  b,  ca,\;b,\;c are in AP and a2,  b2,  c2{a^2},\;{b^2},\;{c^2} are in GP. If a<b<ca < b < c and a+b+c=32a + b + c = \frac{3}{2}, then the value of aa is
View answer and explanation
Correct option: D
12−12\frac{1}{2} - \frac{1}{{\sqrt 2 }}
Option D (12−12\frac{1}{2} - \frac{1}{{\sqrt 2 }}) is correct because solving the given conditions for AP and GP leads to this value for a.
33 Hard
33 of 37
Application +4 | -1
The sum of first nn terms of the series 11+12+14+21+22+24+31+32+34+…\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\frac{3}{1+3^2+3^4}+\ldots is
View answer and explanation
Correct option: B
n(n+1)2(n2+n+1)\frac{n(n+1)}{2\left(n^2+n+1\right)}
Option B (n(n+1)2(n2+n+1)\frac{{n(n+1)}}{{2(n^2+n+1)}}) is correct because it represents the sum of the first n terms of the given series.
34 Hard
34 of 37
Analyzing +4 | -1
The sum of all the products of the first nn natural numbers taken two at a time is
View answer and explanation
Correct option: A
124n(n−1)(n+1)(3n+2)\frac{1}{{24}}n\left( {n - 1} \right)\left( {n + 1} \right)\left( {3n + 2} \right)
Option A (124n(n−1)(n+1)(3n+2)\frac{1}{{24}}n(n - 1)(n + 1)(3n + 2)) is correct because the sum of all products of the first n natural numbers taken two at a time is 124n(n−1)(n+1)(3n+2)\frac{1}{{24}}n(n - 1)(n + 1)(3n + 2).
35 Hard
35 of 37
Application +4 | -1
The sum of first 20 terms of 3⋅5+7⋅52+11⋅53+…3 \cdot 5+7 \cdot 5^2+11 \cdot 5^3+\ldots is
View answer and explanation
Correct option: C
52(1+39⋅520)\frac{5}{2}\left(1+39 \cdot 5^{20}\right)
Option C (52(1+39⋅520)\frac{{5}}{{2}}(1 + 39 \cdot 5^{{20}})) is correct because it represents the sum of the arithmetico-geometric series after applying the appropriate formula and simplifying.
36 Hard
36 of 37
Analyzing +4 | -1
If a,  b,  c,  d  and  pa,\;b,\;c,\;d\;\mathrm{and}\;p are different real numbers such that (a2+b2+c2)  p2−2(ab+bc+cd)    p+(b2+c2+d2)≤0,then  a,  b,  c,  d\left( {{a^2} + {b^2} + {c^2}} \right)\;{p^2} - 2\left( {ab + bc + cd} \right)\;\;p + \left( {{b^2} + {c^2} + {d^2}} \right) \le 0,\mathrm{then}\;a,\;b,\;c,\;d are in
View answer and explanation
Correct option: B
GP
Option B (GP) is correct because the given condition implies that a,b,c,da, b, c, d are in GP, as derived from the inequality.
37 Hard
37 of 37
Application +4 | -1
Let S1,  S2,…{S_1},\;{S_2}, \ldots be squares such that for each n≥1,n \ge 1, the length of a side of Sn{S_n} equals the length of a diagonal of Sn+1.{S_{n + 1}}. If the length of a side of S1{S_1} is 10cm, then for which of the following values of nn is the area of Sn{S_n} less than 1 sq cm?
View answer and explanation
Correct option: C
9
Option C (9) is correct because after determining the side length of successive squares, the area of S_n becomes less than 1 sq cm when n = 9.
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