Find the sum to n terms of series: 5+11+19+29+41.............It is G.P of 11th Std not A.P.
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n(n+1)(2n+1/3+1)-1
n(n+1)(2n+4)/6-1
n(n+1)(2n+4)/6-1
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rajendrapatel25:
Thank you so much
Answered by
30
Answer:
\frac{n(n+1)(2n+4)}{6}-1
Step-by-step explanation:
In the given series,
We have the series,
5 + 11 + 19 + 29 + 41 + .................
So,
We can simply write it as,
5 + 11 + 19 + 29 + 41 + ................. = (2² + 1) + (3² + 2) + (4² + 3) + (5² + 4) + ............
On rearranging, we get,
= (2² + 3² + 4² + 5² +.......) + (1 + 2 + 3 + ........)
Now, we know that sum of squares of first 'n' terms is given by,
So,
We can write the first part as,
also,
Sum of the second half is given by,
Therefore, the total sum of the 'n' terms is given by,
Therefore, the sum of the the terms is given by,
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