2+12+36+80+...nterms=?
nn+1n+23n+524
nn+1n+23n+112
nn+1n+2n+524
nn+1n+2n+312
Explanation for correct option:
Step-1: Simplify the given series
Let the given series, S=2+12+36+80+...n
⇒S=12+13+22+23+32+33+......n⇒S=12+22+32+.....n+13+23+33+.....n
Step-2: Apply formula for sum of first n natural number and their cubes
∵(12+22+32+......n)=n(n+1)(2n+1)6and(13+23+33+......n)=(n(n+1)2)2
⇒S=nn+12n+16+nn+122⇒S=nn+12n+16+n2n+124⇒S=nn+12n+16+nn+14⇒S=nn+122n+1+3nn+112⇒S=nn+14n+2+3n2+3n12⇒S=nn+13n2+7n+212⇒S=nn+13n2+6n+n+212⇒S=nn+13n+1n+212⇒S=nn+1n+23n+112
Therefore, correct answer is option B