श्रेणी के n पदों का योग ज्ञात कीजिए।
Answers
(n/3)(n² + 3n + 5) = 3 + 7 + 13 + 21 + 31 + ....................
Step-by-step explanation:
3 + 7 + 13 + 21 + 31 + ....................
= (3 + 0) + (3 + 4) + (3 + 10) + (3 + 18) + (3 + 28) + ..........
= 3n + (0 + 4 + 10 + 18 + 28 +............................)
= 3n + ( 0*3 + 1 * 4 + 2 *5 + 3*6 + 4*7 +......................)
=> aₙ = (n-1)(n+2)
= 3n + ∑aₙ
= 3n + ( ∑(n-1)(n + 2))
= 3n + ∑ (n² + n - 2)
= 3n + ∑n² + ∑n - ∑ 2
= 3n + n(n+1)(2n+1)/6 + n(n+1)/2 - 2n
= n(n+1)(2n+1)/6 + n(n+1)/2 + n
= (n/6) ( (n+1)(2n+1) + 3(n+1) + 6)
= (n/6) ( 2n² + 3n + 1 + 3n+ 3 + 6)
= (n/6) ( 2n² + 6n + 10)
= (n/6) ( 2n² + 6n + 10)
= (n/3)(n² + 3n + 5)
3 + 7 + 13 + 21 + 31 + .................... = (n/3)(n² + 3n + 5)
और पढ़ें
x के किस मान के लिए संख्याएँ - \dfrac{2}{7},\,x,\,-\dfrac{7}{2} गुणोत्तर श्रेणी में हैं
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मान ज्ञात कीजिए \sum_{k=1}^{11} (2 + 3^k))
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