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Question: Find the value of \(3+5+7+......\text{upto }n\) terms A.\(n\left( n-2 \right)\) B.\({{\left( n+...

Find the value of 3+5+7+......upto n3+5+7+......\text{upto }n terms
A.n(n2)n\left( n-2 \right)
B.(n+2)2{{\left( n+2 \right)}^{2}}
C.n(n+2)n\left( n+2 \right)
D.n2{{n}^{2}}

Explanation

Solution

Given series in in the form of A.P. We have to use the formula of Arithmetic Progression to solve this series. These are nn terms in the series and the answer will be in terms of nn .

Formula used:
Sn=n2[2a+(n1)d]{{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]
Where
aa is First term
dd is common difference
nn Number of terms

Complete step-by-step answer:
3+5+7.....+3+5+7.....+ upto nn terms
Here
a=3a=3 (i.e. the First term)
d=53=2d=5-3=2 (d=d= Common difference)
(Difference between any two consecutive terms)
nn is number of terms
Sn=n2[2a+(π1)d] Sn=n2[2×3+(n1)2] Sn=n2[6+2n2] Sn=n2[2n+4] Sn=n2×2[2n+2] \begin{aligned} & Sn=\dfrac{n}{2}\left[ 2a+\left( \pi -1 \right)d \right] \\\ &\Rightarrow Sn=\dfrac{n}{2}\left[ 2\times 3+\left( n-1 \right)2 \right] \\\ &\Rightarrow Sn =\dfrac{n}{2}\left[ 6+2n-2 \right] \\\ &\Rightarrow Sn=\dfrac{n}{2}\left[ 2n+4 \right] \\\ & \Rightarrow Sn =\dfrac{n}{2}\times 2\left[ 2n+2 \right] \\\ \end{aligned}
Sn=n(n+2)\therefore Sn=n\left( n+2 \right)
So option C is correct answer for given series

Additional information:
If we know that last term then we can also use another formula for the series.
i.e. Sn=n2[a+l]{{S}_{n}}=\dfrac{n}{2}\left[ a+l \right] where ll is the last term
l=a+(n1)dl=a+\left( n-1 \right)d
aa Is First term
nn Number of terms
dd Is common difference

Note: In these types of questions there are some patterns so that there may be Arithmetic Progression or Geometric progress. In this particular question we have Arithmetic Progress and we use the formula for the sum of the term of an Arithmetic Progression.