find the sum to n terms of the series 5+14+19+29+41+.....
Answers
5+14+19+29+41+..... = n(n+2)(n+4)/3
Step-by-step explanation:
Correct Question would be
5 + 11 + 19 + 29 + 41 +.........................
= (2² + 1 ) + (3² + 2) + (4² + 3) + (5² + 4) + (6²+ 5) +.................((n + 1)² + n)
= (2² + 3² + 4² + 5² +..................................+ (n + 1)²) + (1 + 2 + 3 +.........+ n)
= (1² + 2² + 3² + 4² + 5² +..................................+ (n + 1)²) ) - 1 + n(n+1)/2
∑r² = r(r + 1)(2r + 1)/6
r = n + 1
= (n + 1)(n + 2)(2n + 3)/6 - 1 + n(n+1)/2
= ((n + 1)(n + 2)(2n + 3) - 6 + 3n(n+1) )/6
= ((n² + 3n + 2)(2n + 3) - 6 + 3n(n+1) )/6
= (2n³ + 9n² + 13n + 6 - 6 + 3n² + 3n)/6
= (2n³ + 12n² + 16n)/6
= 2n(n² + 6n + 8)/6
= n(n² + 6n + 8)/3
= n(n² + 4n + 2n + 8)/3
= n(n+2)(n+4)/3
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find the sum to n terms of the series
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