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97. Find the sum of the first n terms of the series: 3+ 7 +13 +21 +31 +…

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    Answered by

    Payal Gupta | Contributor-Level 10

    6 months ago

    97. The given series is 3+7+13+21+31+ ……. upto n terms

    So, Sum, Sn = 3+7+13+21+31+ ………….+ an-1 + an

    Now, taking,

    Sn = Sn = [ 3 + 7 + 13 + 21 + 31 + ………. an-1 + an ] [ 3+ 7 + 21 + 31 + … an-1 + an]

    0 = [ 3+ (7 - 3) + (13 - 7) + (13 - 7) + (21 - 13) + …….+ (an an-1) - an]

    0 = 3 + [ 4 + 6 + 8 +…….+ (n-1) terms] an

    ⇒an=3+{n−12[2×4+[(n−1)−1]2]} { ? 4 + 6 + 8 + ……. is sum of A.P. of n 1 terms with a = 4, d = 6- 4 = 2}

    ⇒an=3+{n−12[8+(n−2)2]}

    ⇒an=3+{n−12×2[4+n−2]}=3+(n−1)(n+2)

    an = 3 + n2 + (1 + 2) n + (-1) (2) { ? (a + b) (a + b) = a2 + (b + c) a + bc }

    an = 3 + n2 + n- 2 = n2 + n +1

    sum of series, Sn=∑an=∑n2+∑n+∑1

    =n(n+1)(2n+1)6+n(x+1)2+n

    =n[(n+1)(2n+1)6+n+12+1]

    =n[(n+1)(2n+1)+3(n+1)+66]

    =n[2n2+n+2n+1+3n+3+66]

    =n[2n2+6n+10]6]

    =n[2(n2+3n+5)6]

    =n(n2+3n+5)3

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alok kumar singh

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alok kumar singh

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