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Sequences and Series Test - 49

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Sequences and Series Test - 49
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  • Question 1
    1 / -0
    Find the sum of first 1515 terms of the sequence whose nth{n}^{th} term is 3+4n3+4n.
    Solution
    nthn^{th} term = 3+4 n
    1st1^{st} term = 3+4(1)=7
    2nd2^{nd} term = 3+4(2)=11
    3rd3^{rd} term 3+4(3)=15
    It is an AP with a=7, d=4
    S15=152[2(7)+14(4)]S_{15}=\frac{15}{2}[2(7)+14(4)]
    =15[7+28]= 15[7+28]
    =525= 525

  • Question 2
    1 / -0
    Let r=1nr4=f(n),\sum _{ r=1 }^{ n }{ { r }^{ 4 } } =f\left( n \right) , then r=1n(2r1) 4\sum _{ r=1 }^{ n }{ { \left( 2r-1 \right)  }^{ 4 } } is equal to
    Solution

  • Question 3
    1 / -0
    solve

    Solution
    Pattens is as follows:-

    (209)2=121, (20 - 9)^{2} = 121 ,                  (2411)2=169(24 - 11)^{2} = 169

    So (2813)2=152=225 (28 - 13)^{2} = 15^{2} = 225

    Hence missing place = 225 
  • Question 4
    1 / -0
    r=1nrr4+r2+1 \displaystyle\sum _{ r=1 }^{ n }{ \frac { r }{ { r }^{ 4 }+{ r }^{ 2 }+1 }  } is equal to 
    Solution

  • Question 5
    1 / -0
    C1+2C2+3C3+......+nCnC_1 + 2C_2 + 3C_3 + ...... + nC_n is equal to
    Solution

    Given that,


    C1+2.C2+3.C3+.............nCn {{C}_{1}}+2.{{C}_{2}}+3.{{C}_{3}}+.............n{{C}_{n}}

    =n+2×n(n1)2!+3×n(n2)(n3)3!.......n×1 =n+2\times \dfrac{n\left( n-1 \right)}{2!}+3\times \dfrac{n\left( n-2 \right)\left( n-3 \right)}{3!}.......n\times 1

    =n+n(n1)1+n(n2)(n3)2.......1 =n+\dfrac{n\left( n-1 \right)}{1}+\dfrac{n\left( n-2 \right)\left( n-3 \right)}{2}.......1

    =n[1+(n1)1+(n2)(n3)2.......1 =n[1+\dfrac{\left( n-1 \right)}{1}+\dfrac{\left( n-2 \right)\left( n-3 \right)}{2}.......1

    Put, n-1=N

    =n[1+N1+(N+1)(N1)2.......1 =n[1+\dfrac{N}{1}+\dfrac{\left( N+1 \right)\left( N-1 \right)}{2}.......1

    =n(NC0+NC1+NC2...............NCN) =n\left( {}^{N}{{C}_{0}}+{}^{N}{{C}_{1}}+{}^{N}{{C}_{2}}...............{}^{N}{{C}_{N}} \right)

    =n.2N =n{{.2}^{N}}

    =n.2n1 =n{{.2}^{n-1}}

     

    Hence, this is the answer.

  • Question 6
    1 / -0
    If in a series tn=n+1(n+2)!t_n=\dfrac{n+1}{(n+2)!} then n=010tn\displaystyle\sum^{10}_{n=0}t_n is equal to?
    Solution
     Solution - It is given that tn=n+1(n+2)!tn=n+1+11(n+2)!=(n+2)1(n+2)! th =n+2(n+2)!1(n+2)! \begin{array}{l} \text { Solution - It is given that } \\ \qquad \begin{aligned} t_{n} &=\frac{n+1}{(n+2) !} \\ t n &=\frac{n+1+1-1}{(n+2) !}=\frac{(n+2)-1}{(n+2) !} \\ & \text { th }=\frac{n+2}{(n+2) !}-\frac{1}{(n+2) !} \end{aligned} \end{array}

    tn=1(n+1)!1(n+2)! t n=\frac{1}{(n+1) !}-\frac{1}{(n+2) !} n=010tn=n=010(1(n+1)!1(n+2)!) \sum_{n=0}^{10} t_{n}=\sum_{n=0}^{10}\left(\frac{1}{(n+1) !}-\frac{1}{(n+2) !}\right)


    =11(12!+12!13!+13!14!111!+111!)112!=1112! Hence, (C) is the correct option.  \begin{aligned} &=\frac{1}{1}\left(-\frac{1}{2 !}+\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{3 !}-\frac{1}{4 !} \cdots \cdots-\frac{1}{11 !}+\frac{1}{11 !}\right)-\frac{1}{12 !} \\ &=1-\frac{1}{12 !} \\ \text { Hence, } &(\mathrm{C}) \text { is the correct option. } \end{aligned}
  • Question 7
    1 / -0
    insert the missing number:  5,8,12,17,23,___,38 
    Solution
    5,8,12,17,23,......385,8,12,17,23,......38
    5+3=8;8+4=12,12+5=17,17+6=235+3=8;8+4=12,12+5=17,17+6=23
    23+7=30,30+8=3023+7=30,30+8=30
    30\therefore 30
  • Question 8
    1 / -0
    There is a certain relationship between the pair of figure on either side of ::. Identify the relationship on the left side and find the missing figure.

    Solution
    As per the first relation, in the first figure, there is a square on which there is 3 circle and in the second figure, there is a circle on which there is 33 square.

    Following the same in second relation, the correct option is BB.
  • Question 9
    1 / -0
    Rahul told Anand, "Yesterday I defeated the only brother of the daughter of my grandmother." Whom did Rahul defeat ?
    Solution
    Daughter of Grandmother \rightarrow Aunt Aunt's only brother \rightarrow Father
  • Question 10
    1 / -0
    Which number replaces the question mark ?

    Solution
    In each triangle, the central value equals the average of the 3 values around the outside.

    Hence, the required value = 15+17+43=12\dfrac{15+17+4}{3}=12

    Therefore, option B is correct.
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